College Algebra - Solving Exponential Functions

Solve the function: $2^x = 4^{3x+1}$

$-\frac{2}{5}$

$-\frac{4}{5}$

$\frac{1}{2}$

$-\frac{1}{2}$

$\frac{5}{3}$

Explanation

Change the base of the right side to base two.

Rewrite the equation.

$2^x = 2^{2(3x+1)}$

With similar bases, set the powers equal.

Solve for x.

Subtract from both sides.

Divide by negative five on both sides.

$\frac{-5x}{-5} =\frac{ 2}{-5}$

The answer is: $-\frac{2}{5}$

Solve for .

$4^{2x}=4^{-2x-1}$

$x=-\frac{1}{4}$

$x=\frac{2}{3}$

$x=-\frac{3}{8}$

Explanation

Since the bases of the exponents are the same (4), the exponents have to equal each other for the exponential equation to be true.

$x=-\frac{1}{4}$

Solve: $3^{2x+1} =27$

$\frac{9}{2}$

$\frac{3}{2}$

Explanation

Change the right side to base three.

$3^{2x+1} =3^3$

With similar bases, the exponents can be set equal.

Subtract one from both sides.

The answer is:

Solve for x:

$5^{x}=125$

Explanation

In order to solve for x when it is in the exponent, we need to make the bases on either side of the equation look the same. In order to do this, factor the larger number until it looks like the smaller one. For this problem that looks as follows:

$5^{x}=125$

Factor the larger number:

$125=25\cdot5=5^{2}\cdot5=5^{3}$

Now substitute in the factored number:

$5^{x}=5^{3}$

Since the bases are now the same:

Solve for x:

$4^{x}=64$

Explanation

$4^{x}=64$

Factor the larger number:

$64=4\cdot16\rightarrow4\cdot 4^{2}=4^{3}$

Now substitute in the factored number:

$4^{x}=4^{3}$

Since the bases are now the same:

Solve for x:

$2^{x}=8$

$x=\frac{1}{2}$

Explanation

$2^{x}=8$

Factor the larger number:

$8=2^{3}$

Now substitute in the factored number:

$2^{x}=2^{3}$

Since the bases are now the same:

Solve: $4e^{3x}-4=2$

$\frac{3}{2}\ln 2$

$\frac{1}{3}\ln\frac{3}{4}$

None of these

Explanation

$4e^{3x}-4=2$

add four to both sides:

$4e^{3x}=6$

divide both sides by four:

$e^{3x}=\frac{6}{4}$

take the natural logarithm of both sides:

$\ln e^{3x}=\ln\frac{6}{4}$

$3x=\ln\frac{3}{2}$

Divide both sides by three:

$x=\frac{\ln \frac{3}{2}}{3}\approx \boldsymbol{.135}$

Solve for x: $2^{2x}=128$

$x=\frac {7}{2}$

$x=\frac {5}{2}$

Explanation

Step 1: We will take log of base 2 of both sides of the equation. The log base 2 will cancel out the base of the exponential term and just leave the exponent.

$log_2(2^{2x})=log_2(128)$

Step 2: We need to find the value of the right side. To find this number, we do the following:

Let be the solution to the right hand side:

Apply the rule: Raise the base of the log to the power (the solution), and this expression is equal to the value next to the base of the exponent.

So:

If $128\equiv 2^7$ , we can rewrite the equation above:

So, .

Since the bases of the two exponential functions are the same, then the exponents of both exponential functions must also be the same.

The answer is:

Give the solution set for the exponential equation shown below for the following cases:

Case 1,

Case 2

$a^{10x-1}=b^{10x-1}$

Case 1:

All real values of are solutions:

$x=\left {x| -\infty < x <\infty \right }$

Case 2:

$x = \left { \frac{1}{10}\right }$

Case 1:

There are no solutions,

$x =\left { \varnothing \right }$

Case 2:

$x = \left { 0\right }$

Case 1:

$x = \left { \frac{1}{10}\right }$

Case 2:

$x = \left { \frac{1}{10}\right }$

Case 1:

$x=\left {x| -\infty < x <\infty \right }$

Case 2:

There is not enough information to find if we do not know the values of and b.

Case 1:

All real values of are solutions:

$x=\left {x| -\infty < x <\infty \right }$

Case 2:

All real values of are solutions:

$x=\left {x| -\infty < x <\infty \right }$

Explanation

Case 1,

$a^{10x-1}=b^{10x-1}$

$a^{10x-1}=a^{10x-1}$

This is true for all real values of , therefore,

$x=\left {x| -\infty < x <\infty \right }$

Case 2

$a^{10x-1}=b^{10x-1}$

Take the natural logarithm of both sides (note that we could use any logarithm but it's convenient to just choose the natural logarithm).

$\ln\left(a^{10x-1} \right ) = \ln\left[b^{10x-1}\right]$

Use the rule for pulling out exponents $\ln(u^c)=c\ln(u)$ .

$(10x-1)\ln(a) = (10x-1)\ln(b)$

Note that if you were to divide out by , you would obtain $\ln(a)=\ln(b)$ which would imply which is not true for this case. We are solving for .

Expand with the distributive property,

$10x\ln(a)-\ln(a)=10x\ln(b)-\ln(b)$

Collect and isolate terms with onto one side of the equation,

$10x\ln(a)-10x\ln(b)=\ln(a)-\ln(b)$

Factor out (you could just factor out and leave the in front of the logarithms but it's easier to see the solution writing it this way).

$10x[\ln(a)-\ln(b)]=\ln(a)-\ln(b)$

$10x = \frac{\ln(a)-\ln(b)}{\ln(a)-\ln(b)}$

$x = \frac{1}{10}$

What's remarkable about this solution is that it was obtained without specifying any value for or . The solution is true so as long as .

Solve for x:

$2^{x}=32$

Explanation

$2^{x}=32$

Factor the larger number:

$32=4\cdot 8=2^{2}\cdot 2^{3}=2^{5}$

Now substitute in the factored number:

$2^{x}=2^{5}$

Since the bases are now the same:

Solving Exponential Functions

Help Questions

Explanation

Explanation

Explanation

Explanation

Explanation

Explanation

Explanation

Explanation

Explanation

Explanation