Indices With The Same Unknown

Solution w/ quite detail explanation

$5^x \times 5^x \times 3^{x-1} \times 3^{2x}= 125^x \times 3^{2x} \times 3^{2x}$

Rewrite the equation by simplifying it. (optional)

$5^{2x} \times 3^{x-1} \times 3^{2x} = 5^{3x} \times 3^{2x} \times 3^{2x}$

Divide both sides by $3^{2x}$ since both sides have the factor of it.

$5^{2x} \times 3^{x-1} = 5^{3x} \times 3^{2x}$

Divide $5^{2x}$ and $3^{2x}$ both sides.

$\frac{3^{x-1}}{3^{2x}}$ = $\frac{5^{3x}}{5^{2x}}$

Now simplify both sides.

$3^{x-1-2x} = 5^{3x-2x}$

$3^{-x-1} = 5^x$

$3^{-x-1}$ can actually be written as $3^{-x} \times 3^{1}$

Therefore,

$3^{-x} \times 3^{-1} = 5^x$

Divide $3^{-x}$ both sides to get $15^x$

$3^{-1}=\frac{5^{x}}{3^{-x}}$

$\frac{1}{3}=\frac {5^{x}}{\frac{1}{3^{x}}}$

Note that $3^{-1}$ can be written as $\frac{1}{3}$

$\frac{1}{3}=5^{x} \times 3^x$

$15^x=\frac{1}{3}$

Hence,

$3(15^x)=3(\frac{1}{3})$

$3(15^x)=1$

$\begin{aligned} 5^x \times 5^x \times 3^{x-1} \times 3^{2x} & = 125^x \times 3^{2x} \times 3^{2x} \\ 5^{x+x}3^{x-1+2x} & = (5^3)^x 3^{2x+2x} \\ 5^{2x}3^{3x-1} & = 5^{3x} 3^{4x} \\ 5^{2x \color{#3D99F6}{-2x}}3^{3x-1 \color{#D61F06}{-3x+1}} & = 5^{3x \color{#3D99F6}{-2x}} 3^{4x \color{#D61F06}{-3x+1}} \\ 5^0 3^0 & = 5^x 3^{x+1} \\ \Rightarrow 3(3^x)(5^x) & = 1 \\ 3 (3 \times 5)^x & = 1 \\ \Rightarrow 3(15^x) & = \boxed{1} \end{aligned}$

Just by knowing $a^{m} • a^{n} = a^{m+n}$ $a^{m} • b^{m} = (ab)^{m}$ Anyone can do it!!

$5^{x} \times 5^{x} \times 3^{x-1} \times 3^{2x} = 125^{x} \times 3^{2x} \times 3^{2x}$

$5^{x+x} \times 3^{x-1+2x} = (5^{3})^{x} \times 3^{2x+2x}$

$5^{2x} \times 3^{3x-1} = 5^{3x} \times 3^{4x}$

$1= \frac{5^{3x} \times 3^{4x}}{5^{2x} \times 3^{3x-1}}$

$1= 5^{3x-2x} \times 3^{4x-(3x-1)}$

$1=5^{x} \times 3^{x+1}$

$1=5^{x} \times 3^{1} \times 3^{x}$

$1= 3(15^{x})$