Solving Exponential & Logarithmic Equations - Algebra 3

1. The two sides of $\log_4 x = \log_2(x-2)$ use different bases. Convert to a common base and solve.
   Answer: $x=4$ Since $\log_4 x = \tfrac{1}{2}\log_2 x$, the equation becomes $\log_2 x = \log_2 (x-2)^2$, so $x=(x-2)^2$ and $x^2-5x+4=0$ with roots $x=4$ and $x=1$. The domain requires $x>2$, so $x=1$ is extraneous because it makes $\log_2(-1)$ undefined, while $x=4$ gives $\log_4 4 = \log_2 2 = 1$. Keeping both roots skips that check, and a valid solution clearly exists.
2. Treat $\log_3 x$ as a single variable to solve $(\log_3 x)^2 - 4\log_3 x + 3 = 0$.
   Answer: $x=3$ and $x=27$ Letting $u=\log_3 x$ gives $u^2-4u+3=0$, so $u=1$ or $u=3$, and undoing the substitution gives $x=3^1=3$ and $x=3^3=27$; both are positive, so neither is extraneous. Reporting $1$ and $3$ stops at the $u$ values. Reporting $9$ and $27$ uses $u=2$ in place of $u=1$, and keeping only $x=3$ discards a valid root.
3. For which values of $x$ is the expression $\log_5(x^2-4x)$ defined?
   Answer: $x<0$ or $x>4$ The argument must be positive, so $x(x-4)>0$, which holds when both factors are negative or both are positive: $x<0$ or $x>4$. The condition $x>4$ captures only half of that set and omits values such as $x=-1$, where the argument is $5$. On $0<x<4$ the product is negative, as at $x=2$, where the argument is $-4$. Excluding only $0$ and $4$ still admits that entire interval.
4. Find every real $x$ satisfying $3^{2x} - 10\cdot 3^x + 9 = 0$.
   Answer: $x=0$ and $x=2$ Substituting $u=3^x$ gives $u^2-10u+9=0$, so $u=1$ or $u=9$. Both are positive, so both survive: $3^x=1$ gives $x=0$ and $3^x=9$ gives $x=2$. Reporting $1$ and $9$ leaves the answer in terms of $u$. Discarding $u=1$ wrongly treats it as extraneous, since $3^0=1$ is perfectly valid.
5. A student solves $\log_2(3x) = \log_2 x + \log_2(x-2)$ and finds $x=0$ and $x=5$. Which values actually solve the equation?
   Answer: Only $x=5$ solves it. Condensing the right side gives $3x=x(x-2)$, so $x^2-5x=0$ and the roots are $0$ and $5$. The domain requires $x>2$, and substituting $x=0$ gives $\log_2 0$, which is undefined, so that root is extraneous even though it is not negative. Substituting $x=5$ gives $\log_2 15 = \log_2 5 + \log_2 3$, so exactly one value survives rather than both or neither.
6. Solve $(\log_2 x)^2 = \log_2(x^4)$, taking care with every candidate value of $\log_2 x$.
   Answer: $x=1$ and $x=16$ With $u=\log_2 x$ the equation is $u^2=4u$, so $u=0$ or $u=4$, giving $x=2^0=1$ and $x=2^4=16$. Both are positive, so neither is extraneous: $u=0$ is a legitimate logarithm value even though $0$ is never a legitimate argument. Discarding $u=0$ confuses those two roles and loses $x=1$, while reporting $x=0$ treats a $u$ value as an $x$ value.
7. The equation $3^{x+1} = 5^{x}$ has unlike bases. Which expression is its exact solution?
   Answer: $x=\frac{\ln 3}{\ln 5 - \ln 3}$ Taking $\ln$ of both sides gives $(x+1)\ln 3 = x\ln 5$, so $\ln 3 = x(\ln 5 - \ln 3)$ and $x=\frac{\ln 3}{\ln 5 - \ln 3}$. The expression $\frac{\ln 3}{\ln 5}$ drops the $x\ln 3$ term when collecting like terms. The expression $\frac{\ln 5}{\ln 3}-1$ divides before isolating $x$. The last expression divides by $\ln 3$ rather than by the difference.
8. How many real solutions does $\log_4 x + \log_4(x+6) = 2$ have?
   Answer: Exactly one Condensing gives $x(x+6)=16$, so $x^2+6x-16=0$ and the roots are $x=2$ and $x=-8$. Only $x=2$ meets the domain requirement $x>0$, so exactly one solution survives. Answering two counts the extraneous root $x=-8$, which makes both logarithms undefined. Answering none discards the valid root $x=2$, and a logarithmic equation of this form cannot have infinitely many solutions.
9. Solve $\log_2(\log_3 x)=2$.
   Answer: $x=81$ When you see nested logarithms, think of unwrapping from the outside in. The equation $\log_2(\log_3 x)=2$ asks: the base-2 logarithm of some number equals $2$, so that inner number must be $2^2=4.$ Thus you have $\log_3 x=4.$ Now convert to exponential form: if $\log_3 x=4$, then $x=3^4=81.$ Why do the other choices miss? Choosing $x=9$ usually comes from stopping at $\log_3 x=2$ and forgetting the outer log; $9=3^2$ would solve the inner equation only. Choosing $x=16$ likely comes from rewriting $\log_3 x=4$ incorrectly as $x=4^2$; but exponentiation must use the base of the logarithm, $3$, not the value $4$. Choosing $x=64$ likely comes from exchanging the roles and computing $4^3=64$ instead of $3^4=81$. The base of the logarithm is the base being raised to the exponent, not the final exponent itself. A reliable habit for nested logs: translate each logarithmic layer into its exponential form immediately — $\log_b y = a$ means $y=b^a$ — and do this one layer at a time, starting with the outermost logarithm. That turns nested-log problems into simple arithmetic.
10. Compare the solution sets of $\log_3(x^2) = 4$ and $2\log_3 x = 4$. Which description is accurate?
    Answer: The first has $\{-9,9\}$ and the second has $\{9\}$. The first equation needs only $x^2>0$, so $x^2=81$ admits $x=9$ and $x=-9$, and substituting $-9$ gives $\log_3 81 = 4$. The second equation contains $\log_3 x$ directly, which requires $x>0$ and leaves only $x=9$. The two sets therefore differ, which is why rewriting $\log_3(x^2)$ as $2\log_3 x$ is valid only for positive $x$.
11. Solve the equation $\ln(2x-1) = 0$ and verify the result lies in the domain.
    Answer: $1$ Since $\ln A = 0$ exactly when $A=1$, the equation gives $2x-1=1$ and $x=1$, which satisfies the requirement $2x-1>0$. The value $\tfrac{1}{2}$ comes from setting the argument equal to $0$, which makes the logarithm undefined. The value $\tfrac{e}{2}$ comes from setting the argument equal to $e$ rather than $1$, and $0$ makes the argument negative.
12. The equation $\log_2 x + \log_2(x+3) = 2$ yields the candidates $x=1$ and $x=-4$. Why is $x=-4$ not a solution?
    Answer: It makes the argument of a logarithm negative, which is undefined. Substituting $x=-4$ gives $\log_2(-4)$ and $\log_2(-1)$, and logarithms are defined only for positive arguments, so the candidate is extraneous. It does satisfy $x^2+3x-4=0$, which is why the algebra produced it in the first place, so the claim that it fails the quadratic is false. The left side is undefined rather than equal to zero. A negative value can solve a logarithmic equation such as $\log_2(x+8)=2$, so the blanket claim about negative numbers is also false.
13. A colony grows according to $P = 500e^{0.04t}$, with $t$ in years. Which expression gives the time when the colony reaches $1500$?
    Answer: $t=\frac{\ln 3}{0.04}$ Dividing by $500$ first gives $e^{0.04t}=3$, so $0.04t=\ln 3$ and $t=\frac{\ln 3}{0.04}$. The expression $\frac{\ln 1500}{0.04}$ takes the logarithm before dividing by the initial amount. The expression $\frac{3}{0.04}$ never applies a logarithm. The expression $0.04\ln 3$ multiplies by the rate instead of dividing by it.
14. A student solves $\log(x-1) + \log(x-4) = 1$, obtains $x=6$ and $x=-1$, and reports both. Assess this work.
    Answer: Only $x=6$ is a solution of the original equation. The algebra is sound: $(x-1)(x-4)=10$ gives $x^2-5x-6=0$ with roots $6$ and $-1$. The domain requires $x>4$, and substituting $x=-1$ produces $\log(-2)$, which is undefined, so that root is extraneous. Substituting $x=6$ gives $\log 5 + \log 2 = \log 10 = 1$, so the larger root is genuine and the answer is not that both or neither work.
15. Two conditions hold at once: $2^x \cdot 4^y = 64$ and $\frac{2^x}{2^y} = 8$. Find the pair $(x,y)$.
    Answer: $(4,1)$ Writing both sides with base $2$ gives $2^{x+2y}=2^6$ and $2^{x-y}=2^3$, so $x+2y=6$ and $x-y=3$, which yield $y=1$ and $x=4$; checking, $2^4\cdot 4 = 64$ and $16/2=8$. The pair $(5,1)$ satisfies the second equation but gives $128$ in the first. The pair $(2,2)$ satisfies the first but gives $1$ in the second. The pair $(3,3)$ fails both conditions.
16. Express both sides with base $3$ to solve $\left(\tfrac{1}{3}\right)^x = 81$.
    Answer: $-4$ Since $\tfrac{1}{3}=3^{-1}$ and $81=3^4$, the equation becomes $3^{-x}=3^4$, so $-x=4$ and $x=-4$; checking, $\left(\tfrac{1}{3}\right)^{-4}=3^4=81$. The value $4$ drops the negative exponent hidden in the reciprocal base. The value $-3$ mistakes $81$ for $3^3$. The value $3$ makes both errors at once.
17. What value of $x$ satisfies $\log_5(x+4) - \log_5(x-4) = 1$?
    Answer: $6$ The quotient rule gives $\frac{x+4}{x-4}=5$, so $x+4=5x-20$ and $x=6$, which satisfies the domain requirement $x>4$. The value $24$ comes from solving $4x=24$ but not dividing by $4$. The value $-4$ comes from writing $x+4=5x+20$ with a sign error. The value $4$ makes $\log_5(0)$ undefined.
18. Notice that $4^x=(2^x)^2$ and use it to solve $4^x + 2^x - 6 = 0$.
    Answer: $x=1$ Letting $u=2^x$ gives $u^2+u-6=0$, so $u=2$ or $u=-3$. An exponential with base $2$ is always positive, so $u=-3$ is rejected and $2^x=2$ gives $x=1$. Treating $2$ and $-3$ as the $x$ values forgets to undo the substitution, $\log_2(-3)$ is undefined, and the equation does have a real solution.
19. Solve $\log_2(x-1) + \log_2(x+1) = \log_2(8x-16)$ and state which roots survive the domain check.
    Answer: $x=3$ and $x=5$ Equating arguments gives $x^2-1=8x-16$, so $x^2-8x+15=0$ with roots $x=3$ and $x=5$. The domain requires $x>1$ and $8x-16>0$, that is $x>2$, and both roots clear it: $x=3$ gives $\log_2 2 + \log_2 4 = \log_2 8$ and $x=5$ gives $\log_2 4 + \log_2 6 = \log_2 24$. Discarding either root assumes a rejection that the domain does not actually force.
20. Determine the solution set of $\log_5(x-1) + \log_5(x+3) = \log_5(1-x)$.
    Answer: The solution set is empty Equating arguments gives $x^2+3x-4=0$ with roots $x=-4$ and $x=1$, but the domain demands $x>1$ from the left side and $x<1$ from the right side, conditions that cannot hold together. Substituting $x=1$ yields $\log_5 0$ and substituting $x=-4$ yields $\log_5(-5)$, so both roots are extraneous and no real solution remains.

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