Solve for c 3^(c+6)=4^(2-c)

$3^{c + 6} = 4^{2 - c}$

Step 1

Take the log of both sides of the equation.

$\ln (3^{c + 6}) = \ln (4^{2 - c})$

Step 2

Expand $\ln (3^{c + 6})$ by moving $c + 6$ outside the logarithm.

$(c + 6) \ln (3) = \ln (4^{2 - c})$

Step 3

Expand $\ln (4^{2 - c})$ by moving $2 - c$ outside the logarithm.

$(c + 6) \ln (3) = (2 - c) \ln (4)$

Step 4

Solve the equation for $c$ .

Simplify $(c + 6) \ln (3)$ .

Start simplifying.

$0 + 0 + (c + 6) \ln (3) = (2 - c) \ln (4)$

Simplify by adding zeros.

$(c + 6) \ln (3) = (2 - c) \ln (4)$

Apply the distributive property.

$c \ln (3) + 6 \ln (3) = (2 - c) \ln (4)$

Apply the distributive property.

$c \ln (3) + 6 \ln (3) = 2 \ln (4) - c \ln (4)$

Add $c \ln (4)$ to both sides of the equation.

$c \ln (3) + 6 \ln (3) + c \ln (4) = 2 \ln (4)$

Subtract $6 \ln (3)$ from both sides of the equation.

$c \ln (3) + c \ln (4) = 2 \ln (4) - 6 \ln (3)$

Factor $c$ out of $c \ln (3) + c \ln (4)$ .

Factor $c$ out of $c \ln (3)$ .

$c (\ln (3)) + c \ln (4) = 2 \ln (4) - 6 \ln (3)$

Factor $c$ out of $c \ln (4)$ .

$c (\ln (3)) + c (\ln (4)) = 2 \ln (4) - 6 \ln (3)$

Factor $c$ out of $c (\ln (3)) + c (\ln (4))$ .

$c (\ln (3) + \ln (4)) = 2 \ln (4) - 6 \ln (3)$

Divide each term in $c (\ln (3) + \ln (4)) = 2 \ln (4) - 6 \ln (3)$ by $\ln (3) + \ln (4)$ and simplify.

Divide each term in $c (\ln (3) + \ln (4)) = 2 \ln (4) - 6 \ln (3)$ by $\ln (3) + \ln (4)$ .

$\frac{c (\ln (3) + \ln (4))}{\ln (3) + \ln (4)} = \frac{2 \ln (4)}{\ln (3) + \ln (4)} + \frac{- 6 \ln (3)}{\ln (3) + \ln (4)}$

Simplify the left side.

Cancel the common factor of $\ln (3) + \ln (4)$ .

Cancel the common factor.

$\frac{c (\ln (3) + \ln (4))}{\ln (3) + \ln (4)} = \frac{2 \ln (4)}{\ln (3) + \ln (4)} + \frac{- 6 \ln (3)}{\ln (3) + \ln (4)}$

Divide $c$ by $1$ .

$c = \frac{2 \ln (4)}{\ln (3) + \ln (4)} + \frac{- 6 \ln (3)}{\ln (3) + \ln (4)}$

Simplify the right side.

Combine the numerators over the common denominator.

$c = \frac{2 \ln (4) - 6 \ln (3)}{\ln (3) + \ln (4)}$

Factor $2$ out of $2 \ln (4) - 6 \ln (3)$ .

Factor $2$ out of $2 \ln (4)$ .

$c = \frac{2 \ln (4) - 6 \ln (3)}{\ln (3) + \ln (4)}$

Factor $2$ out of $- 6 \ln (3)$ .

$c = \frac{2 \ln (4) + 2 (- 3 \ln (3))}{\ln (3) + \ln (4)}$

Factor $2$ out of $2 \ln (4) + 2 (- 3 \ln (3))$ .

$c = \frac{2 (\ln (4) - 3 \ln (3))}{\ln (3) + \ln (4)}$

Step 5

The result can be shown in multiple forms.

Exact Form:

$c = \frac{2 (\ln (4) - 3 \ln (3))}{\ln (3) + \ln (4)}$

Decimal Form:

$c = - 1.53691286 \dots$

Solve for c 3^(c+6)=4^(2-c)

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