Take the logarithm of both sides with base a: $x = \log_a(b)$. Or, take the natural logarithm: $x = \frac{\ln(b)}{\ln(a)}$

Rewrite the equation in exponential form: $x = b^a$.

1. Use product rule: $\log_b((x+2)x) = c$. 2. Convert to exponential form: $(x+2)x = b^c$. 3. Solve the quadratic equation. 4. Check for extraneous solutions.

1. Since the bases are equal, set the exponents equal to each other: $f(x) = g(x)$. 2. Solve for x. 3. Check for extraneous solutions.

1. Take the logarithm of both sides (either base a or natural log). 2. Solve for x. 3. Check for extraneous solutions.

1. Swap x and y: $x = 2e^{y+1} - 3$. 2. Isolate the exponential term: $(x+3)/2 = e^{y+1}$. 3. Take the natural log: $\ln((x+3)/2) = y+1$. 4. Solve for y: $y = \ln((x+3)/2) - 1$.

1. Swap x and y: $x = \log_3(y-2) + 1$. 2. Isolate the log term: $x-1 = \log_3(y-2)$. 3. Convert to exponential form: $3^{x-1} = y-2$. 4. Solve for y: $y = 3^{x-1} + 2$.

$\log_b(MN) = \log_b(M) + \log_b(N)$

$\log_b(M/N) = \log_b(M) - \log_b(N)$

$\log_b(M^p) = p \cdot \log_b(M)$

$\log_b(M) = \frac{\log_c(M)}{\log_c(b)}$

$f(x) = ab^{x-h} + k$

$f(x) = a \log_b(x-h) + k$

$f^{-1}(x) = h + \frac{\ln((x-k)/a)}{\ln(b)}$

$f^{-1}(x) = b^{\frac{(x-k)}{a}} + h$

Exponential: Isolate exponential term, take logarithm. Logarithmic: Isolate log term, convert to exponential form. Extraneous solutions are more common in logarithmic equations.

Exponential: Domain is all real numbers. Logarithmic: Domain is positive real numbers (argument must be > 0).