Equations Involving Logarithms and Exponentials

Sol A $\log_3(2x-7)=2 \Leftrightarrow 2x-7=3^2=9 \Leftrightarrow 2x=16 \Leftrightarrow x=8$.

Sol B Dividing by 5 gives $e^{7t-4}=6$, and then taking natural logarithms of both sides gives $\ln e^{7t-4}=\ln 6$ or $7t-4=\ln 6$. Therefore $7t=4+\ln6$, so $t=1/7(4+\ln 6)$.

Sol C Factoring $e^{2x}-2e^x-15=0$ gives $(e^x-5)(e^x+3)=0$, so either $e^x=5$ or $e^x=-3$. If $e^x=5$, then $x=\ln5$. Since $e^x>0$ for all values of $x$, $e^x=-3$ has no solution; so $x=\ln5$ is the only solution.

Sol D $\log_2(x+35)-\log_2x=3 \Rightarrow \log_2\frac{x+35}{x}=3$ $\Rightarrow \frac{x+35}{x}=2^3=8 \Rightarrow x+35=8x \Rightarrow 35=7x$ $\Rightarrow x=5$. (Notice that this answer checks in the original equation.)

Sol E $\log_3(5-x)+\log_3(3-x)=\log_3(19-5x) \Rightarrow$ $\log_3((5-x)(3-x))=\log_3(19-5x)\Rightarrow (5-x)(3-x)=19-5x$ $\Rightarrow x^2-8x+15=19-5x \Rightarrow x^2-3x-4=0 \Rightarrow$ $(x-4)(x+1)=0\Rightarrow x=4$ or $x=-1$. Since $x=4$ does not satisfy the original equation, $x=-1$ is the only answer.

Sol 1 $\log_6(x-11)+\log_6(x-6)=2 \Rightarrow \log_6(x-11)(x-6)=2$ $\Rightarrow (x-11)(x-6)=6^2 \Rightarrow x^2-17x+66=36 \Rightarrow$ $x^2-17x+30=0 \Rightarrow (x-15)(x-2)=0 \Rightarrow x=15$ or $x=2$. Since $x=2$ does not check in the original equation, $x=15$ is the only answer.

Sol 2 $(\ln x)^2=\ln(x^4) \Rightarrow (\ln x)^2=4\ln x \Rightarrow$ $(\ln x)^2-4\ln x=0 \Rightarrow (\ln x)(\ln x-4)=0 \Rightarrow$ $\ln x=0$ or $\ln x=4$, so $x=e^0=1$ or $x=e^4$.

Sol 3 $2\ln x-\ln(x+2)=0 \Rightarrow 2\ln x=\ln(x+2)\Rightarrow$ $\ln x^2=\ln(x+2) \Rightarrow x^2=x+2 \Rightarrow x^2-x-2=0$ $\Rightarrow (x-2)(x+1)=0 \Rightarrow x=2$ or $x=-1$. Since $x=-1$ does not check in the original equation, $x=2$ is the only answer.

Sol 4 Factoring $e^{2x}-4e^x+3=0$ gives $(e^x-3)(e^x-1)=0$, so either $e^x=3$ or $e^x=1$. Taking natural logarithms gives $x=\ln 3$ or $x=\ln 1=0$.

Sol 5 $\log_3(x+5)-\log_3(x-7)=2 \Rightarrow \log_3\frac{x+5}{x-7}=2$ $\Rightarrow \frac{x+5}{x-7}=3^2=9 \Rightarrow x+5=9(x-7) \Rightarrow$ $x+5=9x-63 \Rightarrow 68=8x \Rightarrow x=68/8=17/2$.

Sol 6 $\ln(x-3)+\ln(x+1)=\ln(x+7) \Rightarrow \ln(x-3)(x+1)=\ln(x+7)$ $\Rightarrow (x-3)(x+1)=x+7 \Rightarrow x^2-2x-3=x+7 \Rightarrow$ $x^2-3x-10=0 \Rightarrow (x-5)(x+2)=0 \Rightarrow x=5$ or $x=-2$. Since $x=-2$ does not check in the original equation, $x=5$ is the only answer.

Sol 7 Dividing both sides of $3(5^{2x+1})=18(2^{5x-3})$ by 3 gives $5^{2x+1}=6(2^{5x-3})$, and then taking natural logarithms of both sides gives $(2x+1)\ln 5=\ln 6+(5x-3)\ln 2$. Then $2x\ln 5+\ln 5=\ln 6+5x\ln 2 -3\ln 2$, so $(2\ln 5)x-(5\ln 2)x=\ln 6-3\ln 2-\ln5$ and therefore $(2\ln 5-5\ln 2)x=\ln 6-3\ln 2-\ln5$.

Thus $x=\frac{\ln 6-3\ln 2-\ln5}{2\ln 5-5\ln 2}$, which we can also write as $x=\frac{\ln 6-\ln 8-\ln 5}{\ln 25-\ln 32}=\frac{\ln 6-\ln 40}{\ln 25-\ln 32}$ $=\frac{\ln 40-\ln 6}{\ln 32-\ln 25}=\frac{\ln 20/3}{\ln 32/25}$.

Return to the Problems for this Topic.

Return to Precalculus Home Page.