Equations Involving Logarithms and Exponentials

Recall that if $x>0$, then

$\log_{b}x=t \Leftrightarrow b^t=x$.

(Notice that $\log_{b}x$ is undefined if $x\le0$.)

We will denote natural logarithms (logarithms with base $e$) by $\ln x$.

Furthermore, $\log_{b}x=\log_{b}y\Rightarrow x=y$.

Ex 1 Solve the equation $\ln x +\ln(x+2)=\ln(x+6)$.

Sol $\ln x +\ln(x+2)=\ln(x+6) \Rightarrow \ln (x(x+2))=\ln(x+6)$ $\Rightarrow x(x+2)=x+6 \Rightarrow x^2+2x=x+6 \Rightarrow x^2+x-6=0$ $\Rightarrow (x+3)(x-2)=0 \Rightarrow x=-3$ or $x=2$. Since $x=-3$ does not check in the original equation, $x=2$ is the only solution.

Ex 2 Solve the equation $6(5^{2t-9})=24$, and write your answer using natural logarithms.

Sol Dividing by 6 gives $5^{2t-9}=4$, and then taking natural logarithms of both sides gives $\ln(5^{2t-9})=\ln 4$ or $(2t-9)\ln 5=\ln 4$. Then $2t\ln 5-9\ln 5=\ln 4$, so $(2\ln 5)t=9\ln 5+\ln 4$ and therefore $t=\frac{9\ln 5+\ln 4}{2\ln 5}$. Since $\ln 4=2\ln 2$, we can also write the answer as $t=9/2+\frac{\ln 2}{\ln 5}$.

Pr A Solve the equation $\log_3(2x-7)=2$.

Pr B Solve the equation $5e^{7t-4}=30$, writing your answer using natural logarithms.

Pr C Find all solutions of $e^{2x}-2e^x-15=0$.

Pr D Solve the equation $\log_2(x+35)-\log_2x=3$.

Pr E Find all solutions of $\log_3(5-x)+\log_3(3-x)=\log_3(19-5x)$.

Pr 1 Solve the equation $\log_6(x-11)+\log_6(x-6)=2$.

Pr 2 Solve the equation $(\ln x)^2=\ln(x^4)$.

Pr 3 Solve the equation $2\ln x-\ln(x+2)=0$.

Pr 4 Solve the equation $e^{2x}-4e^x+3=0$.

Pr 5 Solve the equation $\log_3(x+5)-\log_3(x-7)=2$.

Pr 6 Solve the equation $\ln(x-3)+\ln(x+1)=\ln(x+7)$.

Pr 7 Solve the equation $3(5^{2x+1})=18(2^{5x-3})$, and write your answer using natural logarithms.

Go to Solutions.

Return to Precalculus Home Page.