Solving Exponential and Logarithmic Equations

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

$27 = 3^3$, so $2x = 3$.

$x = \tfrac{5}{3}$

$2^{3x} = 2^5$, so $3x = 5$.

$x = \tfrac{\ln 25}{\ln 7} \approx 1.654$

Isolate the power: $7^x = 25$. Then $x\ln 7 = \ln 25$.

$t = \tfrac{\ln 5}{0.3} \approx 5.365$

Take $\ln$: $0.3t = \ln 5$.

$x = \tfrac{\ln 5}{\ln 5 - \ln 2} \approx 1.757$

$(x - 1)\ln 5 = x\ln 2$, so $x\ln 5 - x\ln 2 = \ln 5$.

$x = 5$

$2x - 1 = 3^2 = 9$, so $x = 5$. Check: $\log_3 9 = 2$.

$x = 2$

Condense: $\ln \tfrac{x + 1}{x - 1} = \ln 3$, so $x + 1 = 3(x - 1)$ and $x = 2$. Both arguments, $3$ and $1$, are positive.

$x = 2$

$x(x + 2) = 2^3 = 8$, so $x^2 + 2x - 8 = (x - 2)(x + 4) = 0$. $x = -4$ is extraneous, since $\log_2(-4)$ is undefined.

$x = \ln 4 \approx 1.386$

With $u = e^x$: $(u - 4)(u + 1) = 0$. $e^x = -1$ is impossible, since $e^x > 0$, so $e^x = 4$.

Substitute $x = -1$ into the original equation.

$x = -1$ is extraneous: $\log(-1)$ is undefined. The only solution is $x = 10$.

Condensing gives $x(x - 9) = 10$, whose solutions are $10$ and $-1$. The original equation needs $x > 9$, so only $10$ survives. Check: $\log 10 + \log 1 = 1 + 0 = 1$.