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## Answers

## kingavisingh

[latex]\left(\dfrac{1}{4}\right)^{2x-3}=32^{3+x}\\ \left(2^{-2}\right)^{2x-3}=\left(2^5\right)^{3+x}\\ 2^{-4x+6}=2^{15+5x}\\ -4x+6=15+5x\\ 9x=-9\\x=-1 [/latex]

## Nunia1567

[latex]\left(\frac{1}{4}\right)^{2x-3}=32^{3+x}\\\\use:\frac{1}{a}=a^{-1}\\\\\left(4^{-1}\right)^{2x-3}=(2^5)^{3+x}\\\\use\ (a^n)^m=a^{n\cdot m}\\\\4^{-1(2x-3)}=2^{5(3+x)}\\\\4^{-2x+3}=2^{15+5x}\\\\(2^2)^{-2x+3}=2^{15+5x}[/latex]

[latex]2^{2(-2x+3)}=2^{15+5x}\\\\2^{-4x+6}=2^{15+5x}\iff-4x+6=15+5x\ \ \ \ |subtract\ 6\ from\ both\ sides\\\\-4x=9+5x\ \ \ \ \ |subtract\ 5x\ from\ both\ sides\\\\-9x=9\ \ \ \ \ |divide\ both\ sides\ by\ (-9)\\\\\boxed{x=-1}[/latex]