Question

$\log_{5}(7-3x)=\log_{5}(x+1)$
$7-3x=x+1$
$-7 -x -7$
$\frac{-4x}{-4}=\frac{-6}{-4}$
$x=1.5$

Explanation:

Step1: Apply logarithm equality rule

If $\log_b(a) = \log_b(c)$, then $a=c$ (for $a>0, c>0, b>0, b
eq1$).
So $7-3x = x+1$

Step2: Rearrange terms to isolate $x$

Subtract $x$ and subtract $7$ from both sides:
$7-3x -x -7 = x+1 -x -7$
Simplify to get $-4x = -6$

Step3: Solve for $x$

Divide both sides by $-4$:
$\frac{-4x}{-4} = \frac{-6}{-4}$
Simplify to $x = \frac{3}{2} = 1.5$

Step4: Verify domain validity

Check if arguments are positive:
For $7-3x$: $7-3(1.5)=7-4.5=2.5>0$
For $x+1$: $1.5+1=2.5>0$
Both are valid, so $x=1.5$ is a valid solution.

Answer:

$x=1.5$ (or $x=\frac{3}{2}$)