Solve for z 9^(z-2)=38

$9^{z - 2} = 38$

Step 1

Take the natural logarithm of both sides of the equation to remove the variable from the exponent.

$\ln (9^{z - 2}) = \ln (38)$

Step 2

Expand $\ln (9^{z - 2})$ by moving $z - 2$ outside the logarithm.

$(z - 2) \ln (9) = \ln (38)$

Step 3

Simplify the left side.

Apply the distributive property.

$z \ln (9) - 2 \ln (9) = \ln (38)$

Step 4

Move all the terms containing a logarithm to the left side of the equation.

$z \ln (9) - 2 \ln (9) - \ln (38) = 0$

Step 5

Move all terms not containing $z$ to the right side of the equation.

Add $2 \ln (9)$ to both sides of the equation.

$z \ln (9) - \ln (38) = 2 \ln (9)$

Add $\ln (38)$ to both sides of the equation.

$z \ln (9) = 2 \ln (9) + \ln (38)$

Step 6

Divide each term in $z \ln (9) = 2 \ln (9) + \ln (38)$ by $\ln (9)$ and simplify.

Divide each term in $z \ln (9) = 2 \ln (9) + \ln (38)$ by $\ln (9)$ .

$\frac{z \ln (9)}{\ln (9)} = \frac{2 \ln (9)}{\ln (9)} + \frac{\ln (38)}{\ln (9)}$

Simplify the left side.

Cancel the common factor of $\ln (9)$ .

Cancel the common factor.

$\frac{z \ln (9)}{\ln (9)} = \frac{2 \ln (9)}{\ln (9)} + \frac{\ln (38)}{\ln (9)}$

Divide $z$ by $1$ .

$z = \frac{2 \ln (9)}{\ln (9)} + \frac{\ln (38)}{\ln (9)}$

Simplify the right side.

Cancel the common factor of $\ln (9)$ .

Cancel the common factor.

$z = \frac{2 \ln (9)}{\ln (9)} + \frac{\ln (38)}{\ln (9)}$

Divide $2$ by $1$ .

$z = 2 + \frac{\ln (38)}{\ln (9)}$

Step 7

The result can be shown in multiple forms.

Exact Form:

$z = 2 + \frac{\ln (38)}{\ln (9)}$

Decimal Form:

$z = 3.65553680 \dots$