Logarithmic Functions & Graphs - Algebra 3

1. The graph of $f(x)=\log_2 x$ is reflected across the $x$-axis and then shifted 3 units to the right. Which function has the resulting graph?
   Answer: $h(x)=-\log_2(x-3)$ When you see transformations of a function like $f(x)=\log_2 x$, always track two things: the order of operations and where each change acts. Reflections and shifts outside the function (like multiplying by –1 or adding a constant) affect the output; changes inside the argument (like replacing $x$ with $x-3$) affect the input. Here, "reflected across the x-axis" means you multiply the entire output by –1, giving $-\log_2 x$. Then "shifted 3 units to the right" means every input is replaced by $x-3$, so the result is $-\log_2(x-3)$. The choice $-\log_2(x+3)$ would be a reflection across the x-axis but a shift left 3 units, not right. The choice $\log_2(3-x)$ is a reflection across the y-axis (because the negative sign is inside the argument) combined with a right shift, not an x-axis reflection. The choice $-\log_2 x + 3$ reflects across the x-axis but then shifts up 3 units, not right. Only $-\log_2(x-3)$ applies the two transformations in the correct order and direction. A useful memory aid: for horizontal shifts, the sign inside the argument is opposite to the direction—$x-3$ moves right, $x+3$ moves left. And a negative sign outside flips vertically; a negative sign inside flips horizontally. On the exam, write out each transformation step by step, and always check whether the shift is inside or outside the function.
2. Let $f(x)=\log_4(6-2x)+5$. Which statement about the graph of $f$ is correct?
   Answer: Domain $(-\infty,3)$, vertical asymptote $x=3$, range all real numbers. When you see a logarithmic function like $f(x)=\log_4(6-2x)+5$, think about the two conditions that define its graph: the argument of the log must be positive, and the log itself can output any real number. The domain comes from solving $6-2x>0$, which gives $-2x>-6$, so $x<3$. Thus the domain is $(-\infty,3)$, and the vertical asymptote occurs where the argument equals zero: $6-2x=0$, so $x=3$. The graph approaches this asymptote but never touches it. As for range, a logarithm alone has range all real numbers, and adding $5$ only shifts the graph vertically, so the range remains all real numbers. That confirms the correct statement: domain $(-\infty,3)$, vertical asymptote $x=3$, range all real numbers. Now look at the traps. The choice with domain $(-\infty,6)$ and asymptote $x=6$ comes from forgetting the coefficient $-2$ and solving $6-x>0$ incorrectly; the asymptote is tied to where the inside becomes zero, which is $x=3$, not $6$. The choice with domain $(3,\infty)$ reverses the inequality sign, a common error when dividing by a negative number — remember $x<3$, not $x>3$. Finally, the choice with domain $(-\infty,3)$ and asymptote $x=3$ but range $y>5$ confuses a logarithmic function with an exponential or square-root function; logs are not bounded below, so the range is not restricted to values above the vertical shift. Your takeaway: for any logarithmic function, set the argument greater than zero for the domain, set it equal to zero for the vertical asymptote, and remember the range is always all real numbers. Watch the sign when solving inequalities involving a negative coefficient.
3. A student attempts to find the inverse of $f(x)=\log_3(x+2)-4$. The student's work is $x=\log_3(y+2)-4$, then $x+4=\log_3(y+2)$, then $3^x+4=y+2$, then $y=3^x+2$. Which statement correctly identifies the error?
   Answer: The third line should be $3^{x+4}=y+2$, because the entire expression $x+4$ becomes the exponent of 3. The conversion from $x+4=\log_3(y+2)$ to exponential form must use the whole left side as the exponent: $3^{x+4}=y+2$. The student wrote $3^x+4$, which is not equivalent. Subtracting 4 before switching variables is not the error, $3^x-4$ is an invalid conversion, and the fourth line is not fixed by changing only a sign.
4. Let $b>1$. Compared with the graph of $f(x)=\log_b x$, the graph of $g(x)=\log_{1/b} x$ is which of the following?
   Answer: The reflection of $f$ across the $x$-axis, with the same domain and $x$-intercept. When you see a logarithm with a fractional base, your first instinct should be to rewrite it using the exponent property of logarithms. Since $1/b = b^{-1}$, we have $\log_{1/b} x = \log_{b^{-1}} x = -\log_b x$. This means $g(x) = -f(x)$. Multiplying a function by $-1$ flips it vertically across the $x$-axis. The domain of any logarithmic function $\log_a x$ is $(0, \infty)$, regardless of the base $a$, and the $x$-intercept is always at $x=1$ because $\log_a 1 = 0$. Therefore, $g$ has the exact same domain and $x$-intercept as $f$, but its $y$-values are negated. Now look at the other choices. The reflection across the $y$-axis would produce $\log_b (-x)$, which has a domain of $(-\infty, 0)$ and an $x$-intercept at $-1$; that is a different function, not what we have here. The reflection across the line $y=x$ would give the inverse, which is $b^x$, an exponential function, not another logarithm. The choice claiming they are the same graph is tempting because both pass through $(1,0)$, but they are not identical: for $x>1$, $f(x)>0$ while $g(x)<0$ (since $b>1$, $f$ is increasing, $g$ is decreasing). Study tip: Memorize the identity $\log_{1/b} x = -\log_b x$. Recognizing this instantly saves you time and prevents confusion about domain or intercepts—a reciprocal base only causes a vertical reflection, never a change in domain.
5. Starting with $y=\log_3 x$, apply a vertical stretch by factor 4, a reflection across the $x$-axis, and a shift right 2 units, in that order. Which equation represents the result?
   Answer: $y=-4\log_3(x-2)$ The vertical stretch and $x$-axis reflection produce $-4\log_3 x$. A shift right 2 replaces $x$ by $x-2$, giving $-4\log_3(x-2)$. The option with $x+2$ shifts left, the option with a subtracted 2 outside shifts the graph down, and $4\log_3(2-x)$ involves a $y$-axis reflection rather than an $x$-axis reflection.
6. Let $f(x)=\log_4(x-2)+3$. If $g$ is the inverse of $f$, what are the horizontal asymptote and $y$-intercept of $g$?
   Answer: horizontal asymptote $y=2$; $y$-intercept $\frac{129}{64}$ Swapping $x$ and $y$ in $y=\log_4(x-2)+3$ gives $x-3=\log_4(y-2)$, so $g(x)=4^{x-3}+2$. The horizontal asymptote is $y=2$, and $g(0)=4^{-3}+2=\frac{1}{64}+2=\frac{129}{64}$. The incorrect options come from using $x+3$ in the exponent, using $-2$ as the vertical shift, or carrying the original $+3$ into the inverse.
7. Which of the following equations, all with base 3, could define a function whose graph has vertical asymptote $x=2$, contains $(3,0)$, and is decreasing on its domain?
   Answer: $f(x)=-\log_3(x-2)$ Whenever you see a transformed logarithmic function, ask three things: Where does the argument become $0$? That gives the vertical asymptote. What is the domain? That tells you whether a stated point can exist. And what is the sign of the coefficient? With base $3>1$, a negative coefficient makes the function decreasing. For an asymptote at $x=2$, the argument must become zero at $x=2$, so both forms with $2-x$ and $x-2$ are candidates. The choices $\log_3(2-x)$ and $-\log_3(2-x)$ have domain $x<2$, so $x=3$ is not in their domain; they cannot contain $(3,0)$. Also, $-\log_3(2-x)$ is actually increasing on its domain, so it fails the decreasing condition as well. Now consider $\log_3(x-2)$. It has domain $x>2$, asymptote $x=2$, and at $x=3$ gives $\log_3(1)=0$, so it contains $(3,0)$. But its derivative is positive, so it increases. The negative version $-\log_3(x-2)$ flips it: same asymptote and same point because $-\log_3(1)=0$, but its derivative is negative, so it decreases. Thus it matches all conditions. Study tip: for $y=\pm\log_b(x-h)$, the asymptote is $x=h$, and with $b>1$, decreasing requires the negative sign. Always test the given point by setting the argument to $1$.
8. The graph of $f(x)=\log_b x$ is reflected across the $x$-axis and then shifted up 3 units. The resulting graph has $x$-intercept $9$. What is $b$?
   Answer: $\sqrt[3]{9}$ The transformed function is $g(x)=-\log_b x+3$. Its $x$-intercept is 9, so $-\log_b 9+3=0$, or $\log_b 9=3$. Therefore $b^3=9$, so $b=\sqrt[3]{9}$. Treating the $x$-intercept 9 as the base, or applying a sign error to obtain the reciprocal, gives the other options.
9. Let $f(x)=\log_2 x$. The graph of $h(x)=\log_2(4x)$ is identical to the graph of which of the following?
   Answer: $f(x)+2$ Using the product property, $\log_2(4x)=\log_2 4+\log_2 x=2+\log_2 x$. Therefore $h(x)=f(x)+2$, a vertical shift up 2. A horizontal shift $f(x+4)$ adds $\log_2(x+4)$, $f(2x)=1+\log_2 x$, and $2f(x)=2\log_2 x$, none of which match.
10. Let $f(x)=\log_b(x-2)+1$. The inverse of $f$ has $y$-intercept $5$. What is $b$?
    Answer: $\frac13$ The inverse of $f(x)=\log_b(x-2)+1$ is $g(x)=b^{x-1}+2$. Its $y$-intercept is $g(0)=b^{-1}+2$. Setting this equal to 5 gives $b^{-1}=3$, so $b=\frac13$. The value 3 comes from dropping the reciprocal exponent, and 9 would be an unnecessary square of 3.
11. The inverse of $f(x)=5^{2x-1}+3$ is $g$. Which statement about $g$ is true?
    Answer: $g(x)=\frac{1}{2}\log_5(x-3)+\frac{1}{2}$; domain $x>3$; vertical asymptote $x=3$ Swapping $x$ and $y$ in $y=5^{2x-1}+3$ gives $x=5^{2y-1}+3$, so $\log_5(x-3)=2y-1$. Thus $g(x)=\frac{1}{2}\log_5(x-3)+\frac{1}{2}$, with domain $x>3$ and vertical asymptote $x=3$. The option with $x+3$ or $-\frac{1}{2}$ mixes up the signs of the shifts; the option with $2\log_5(x-3)-1$ does not divide by 2 when solving for $y$.
12. The point $(5,1)$ lies on the graph of $f(x)=\log_2(x+3)-2$. If $g$ is the inverse of $f$, which point must lie on the graph of $g$?
    Answer: $(1,5)$ Whenever you see a question about an inverse function, think about the defining relationship: if $(a,b)$ lies on $f$, then $(b,a)$ must lie on $f^{-1}$. An inverse swaps the input and output, which geometrically reflects the graph across the line $y=x$. Since $(5,1)$ lies on $f$, the inverse $g$ must contain the swapped point $(1,5)$. That is exactly the point that must lie on the graph of $g$. The choice $(5,1)$ is just the original point on $f$, not on $g$ — it would mean the function and its inverse share the same coordinate pair, which isn't implied here. The choice $(1,-5)$ might come from incorrectly subtracting the $2$ from the $y$-value or misapplying the logarithm's vertical shift; no swapping of coordinates is involved. The choice $(5,-1)$ also swaps incorrectly, perhaps by negating the $y$-coordinate or confusing the $-2$ outside the logarithm with the output value. None of those reflect the actual inverse rule. A quick strategy for inverse questions: whenever you are given a point on a function and asked for a point on its inverse, simply switch the coordinates. No calculation with the logarithm formula is needed. Watch out for traps that keep the same point or alter the sign — the inverse only swaps the ordered pair, nothing else.
13. Let $f(x)=a\log_2(x+3)$. If $f(5)=8$, what is the horizontal asymptote of $f^{-1}$?
    Answer: $y=-3$ The inverse of a logarithmic function has a horizontal asymptote corresponding to the vertical asymptote of the original function. Since $x+3=0$, $f$ has vertical asymptote $x=-3$, so $f^{-1}$ has horizontal asymptote $y=-3$. The parameter $a$ does not change the asymptote. The option $x=-3$ is the original vertical asymptote, not the inverse's horizontal asymptote.
14. Let $f(x)=\log_3(x-6)$. Which statement about the graph of $f^{-1}$ is correct?
    Answer: Its domain is all real numbers, its range is $(6,\infty)$, and it has horizontal asymptote $y=6$. When you see a question about the graph of an inverse function, remember that the inverse swaps the domain and range of the original function, and its graph is a reflection across the line $y=x$. For $f(x)=\log_3(x-6)$, the logarithm is defined only when $x-6>0$, so the domain of $f$ is $(6,\infty)$, and its range is all real numbers. Therefore, the inverse $f^{-1}$ has domain all real numbers and range $(6,\infty)$. Also, because $f$ has a vertical asymptote at $x=6$, its reflection has a horizontal asymptote at $y=6$. That matches the statement giving domain all real numbers, range $(6,\infty)$, and horizontal asymptote $y=6$. The statement claiming domain $(6,\infty)$, range all real numbers, and vertical asymptote $x=6$ actually describes the original function $f$, not its inverse. The statement with range $(0,\infty)$ and horizontal asymptote $y=0$ confuses the inverse with a basic exponential graph like $y=3^x$, but the correct inverse here is $y=3^x+6$, whose range starts at 6. The statement with domain all real numbers, range $(6,\infty)$, but vertical asymptote $x=6$ gets the domain and range right but misidentifies the asymptote: an inverse of a vertical asymptote is a horizontal asymptote, not another vertical one. A useful habit: whenever you find an inverse, immediately swap the domain and range, and swap "vertical" and "horizontal" in any asymptote description.
15. Consider $f(x)=\log_{1/2}(8x)$. Which statement about its graph is true?
    Answer: It is decreasing, its $x$-intercept is $\frac18$, and its vertical asymptote is $x=0$. Since $8x>0$, the domain is $x>0$, so the vertical asymptote is $x=0$. The base $\frac12$ is less than 1, so the graph is decreasing. Setting $\log_{1/2}(8x)=0$ gives $8x=1$, so the $x$-intercept is $\frac18$. The incorrect options either ignore the factor 8, reverse the monotonicity, or misstate the domain and asymptote.
16. The graph of $g$ is the reflection over the line $y=x$ of the graph of $f(x)=3^{x-2}+1$. Which equation defines $g$?
    Answer: $g(x)=\log_3(x-1)+2$ Whenever you see a reflection over the line $y=x$, you are finding the inverse function. That means you swap $x$ and $y$ and then solve for the new $y$. Start with $y=3^{x-2}+1$, swap to get $x=3^{y-2}+1$. Subtract 1 from both sides: $x-1=3^{y-2}$. Now take the base-3 logarithm of each side: $\log_3(x-1)=y-2$. Finally add 2: $y=\log_3(x-1)+2$. So the correct equation is $g(x)=\log_3(x-1)+2$. Now look at the other choices. The one with $\log_3(x+2)-1$ has the numbers swapped and the signs flipped — it looks like someone mistakenly added 2 inside and subtracted 1 outside, which reverses the original shifts incorrectly. The choice $\log_3(x+1)-2$ adds 1 inside and subtracts 2 outside — again the constants are in the wrong places, confusing the horizontal shift (from the $-2$ in the exponent) with the vertical shift (the $+1$ at the end). The choice $\log_3(x-2)+1$ keeps the original shifts intact but does not invert them; it treats the function as its own inverse, which is not true for exponential and logarithmic pairs. For a quick check, remember that the inverse of $a^{x-h}+k$ is $\log_a(x-k)+h$. The horizontal shift $h$ becomes the vertical shift, and the vertical shift $k$ becomes the horizontal shift inside the log. On the exam, always swap and solve — don't just guess based on the original shifts.
17. Let $f(x)=\log_2(x-3)$ and $g(x)=2^x+3$. Which statement is true?
    Answer: The point $(4,0)$ lies on $f$, so the point $(0,4)$ lies on $g$. Since $f(4)=\log_2(4-3)=0$, the inverse relation must contain the reflected point $(0,4)$. Indeed, $g(0)=2^0+3=4$, confirming that $g$ is the inverse of $f$. The other statements use the wrong reflection axis, keep the same point on the inverse, or use the wrong signs for the asymptotes.
18. Let $g$ be the inverse of $f(x)=5^{x-1}$. What is $g(25)$?
    Answer: $3$ When you see a question about an inverse function, remember that $g(y)$ asks: "What input $x$ to $f$ produces the output $y$?" So $g(25)$ means solving $f(x)=25$. Here, $f(x)=5^{x-1}$, so set $5^{x-1}=25$. Since $25=5^2$, the exponents must match: $x-1=2$, giving $x=3$. Thus $g(25)=3$. Now, why the other choices are traps: choosing $26$ comes from mistakenly thinking $g(25)=f(25)+1$, or swapping input/output without solving. Choosing $2$ happens if you solve $5^x=25$ but forget the $-1$ shift in the exponent—that gives $x=2$, but plugging back yields $f(2)=5$, not $25$. Choosing $5$ might arise from thinking the base $5$ is the answer, or confusing $g(25)$ with evaluating $f$ at $25$, which is enormous, not $5$. Your go-to strategy: to find $g(a)$, always write $f(x)=a$ and solve for $x$. Watch for transformations like the $-1$ inside the exponent—they shift your answer. Practice by rewriting powers as the same base, then equate exponents. This turns inverse problems into simple algebra.
19. Let $f(x)=\log_{1/3}(x-2)$. Which statement best describes the graph of $f$?
    Answer: It has vertical asymptote $x=2$, and $f(x)\to -\infty$ as $x\to\infty$. Whenever you see a logarithmic function, first find the domain by setting the inside greater than $0$. That single step gives you the vertical asymptote. Here $f(x)=\log_{1/3}(x-2)$, so the inside is $x-2$, meaning $x>2$ and the asymptote is $x=2$, not $-2$. Since the base $1/3$ is between $0$ and $1$, this log function is decreasing: as each input grows, the output becomes more negative. So as $x\to\infty$, we have $x-2\to\infty$, and $f(x)\to -\infty$. The statement that combines vertical asymptote $x=2$ with $f(x)\to -\infty$ as $x\to\infty$ is correct. Now look at the distractors. The statement saying there is a horizontal asymptote $y=2$ is a red herring: logarithmic functions do not level off to a horizontal asymptote; they diverge as the input grows. That same choice also claims $f(x)\to -\infty$ as $x\to 2^+$, but near a vertical asymptote with base $<1$, the values actually rush upward to $+\infty$. The statement with vertical asymptote $x=2$ and $f(x)\to +\infty$ as $x\to\infty$ correctly identifies the asymptote but reverses the end behavior, since the decreasing log base sends large inputs down to $-\infty$. Finally, the statement with vertical asymptote $x=-2$ is wrong about the location $x=2$ and repeats the wrong end behavior. Study tip: For any log, set the argument equal to $0$ to find the vertical asymptote. Then compare the base to $1$: base $>1$ means increasing, base $<1$ means decreasing. That pattern tells you the behavior at infinity quickly and reliably.
20. Let $h(x)=-\log_3(x+2)$, and let $g$ be the inverse of $h$. Which statement describes the graph of $g$?
    Answer: Domain $(-\infty,\infty)$, range $(-2,\infty)$, horizontal asymptote $y=-2$ The inverse of $h$ is $g(x)=3^{-x}-2$. An exponential expression $3^{-x}$ has domain all real and range $(0,\infty)$, so subtracting 2 gives range $(-2,\infty)$ and horizontal asymptote $y=-2$. The statement with domain $x>-2$, range all real, and vertical asymptote $x=-2$ describes the original $h$, not its inverse; the remaining options have the wrong range or domain sign.

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