# How do you solve log_7 3+log_7x=log_7 32?

Jul 15, 2016

$x = \frac{32}{3} = 10 \frac{2}{3}$

#### Explanation:

If logs to the same base are being added, then the numbers were multiplied.

${\log}_{7} 3 + {\log}_{7} x = {\log}_{7} 32$

${\log}_{7} \left(3 \times x\right) = {\log}_{7} 32$

$3 x = 32 \text{ if log A = log B. then A = B}$

$x = \frac{32}{3} = 10 \frac{2}{3}$

Jul 15, 2016

$x = \frac{32}{3}$

#### Explanation:

Use the formula:

${\log}_{a} \left(x\right) = \frac{{\log}_{b} \left(x\right)}{{\log}_{b} \left(a\right)}$

So our equation becomes:

$\frac{\log \left(3\right)}{\log \left(7\right)} + \frac{\log \left(x\right)}{\log \left(7\right)} = \frac{\log \left(32\right)}{\log \left(7\right)}$

Multiply both sides by $\log \left(7\right)$ and combine using rules of logs

$\log \left(3 x\right) = \log \left(32\right)$

Cancelling the logs by taking exponents yields

$3 x = 32$

$\implies x = \frac{32}{3}$