Question

1. solve the equation.

\log (x + 6) = \log 7x

x = \square

Explanation:

Step1: Use the property of logarithms

If $\log a = \log b$, then $a = b$ (for the same base and valid domain). So we can set the arguments equal:
$x + 6 = 7x$

Step2: Solve for x

Subtract $x$ from both sides:
$6 = 7x - x$
$6 = 6x$

Step3: Divide both sides by 6

$\frac{6}{6} = \frac{6x}{6}$
$1 = x$

We should also check the domain. For $\log(x + 6)$ and $\log(7x)$ to be defined, $x + 6>0$ and $7x>0$. When $x = 1$, $1 + 6 = 7>0$ and $7\times1 = 7>0$, so it is valid.

Answer:

$1$