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(5/6)^4x=(36/25)^9-x, please help solve for x, and the 9-x is all superscript

Sagot :

[tex]\frac{36}{25}=\frac{6^2}{5^2}=\left(\frac{6}{5}\right)^2=\left(\frac{5}{6}\right)^{-2}\\\\therefore:\\\\\left(\frac{5}{6}\right)^{4x}=\left[\left(\frac{5}{6}\right)^{-2}\right]^{9-x}\\\\\left(\frac{5}{6}\right)^{4x}=\left(\frac{5}{6}\right)^{-2(9-x)}\iff4x=-2(9-x)\\\\4x=-2\cdot9-2\cdot(-x)\\\\4x=-18+2x\ \ \ \ \ |subtract\ 2x\ from\ both\ sides\\\\2x=-18\ \ \ \ \ \ |divide\ both\ sides\ by\ 2\\\\\boxed{x=-9}[/tex]


[tex]Use:\\a^{-n}=\left(\frac{1}{a}\right)^n\\\\\left(a^n\right)^m=a^{n\cdot m}\\\\\left(\frac{a}{b} \right)^n=\frac{a^n}{b^n}[/tex]
ollieboyne
(25/36)^2x=(36/25)^9-x
(36/25)^-2x=(36/25)^9-x

-2x=9-x
-x=9
x=-9
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