Question

solve for X? solve this​

Answer 1

$\huge{ \green{ \mathfrak{ \underline{Answer}}}}$

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$\boxed{\red{\bold{Solving\:powers:}}}$

$\displaystyle{\left( \sqrt[3]{\frac{3}{5}} \right)^{2x+1}= \frac{125}{27}}$

$\displaystyle{\left( \sqrt[3]{\frac{3}{5}} \right)^{2x+1 }= \left( \frac{5}{3} \right)^3}$

$\displaystyle{\left( \frac{3}{5} \right)^{\frac{2x+1}{3}}} = \left( \frac{3}{5} \right)^{-3}$

$\boxed{\red{\bold{Equating \: powers:}}}$

$\displaystyle{\frac{2x+1}{3}} = -3$

$\boxed{\red{\bold{Solving\:the\:equation:}}}$

$2x+1 = -9$

$2x = -10$

$x = -5$

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Answer 2

$\huge\star{\green{\underline{\mathfrak{Answer :-}}}}$

$\leadsto\tt{\bigg(\sqrt[3]{\dfrac{3}{5}}\bigg)^{2x+1}=\dfrac{125}{27}}$

$\leadsto\tt{{{\bigg(\bigg(\sqrt[3]{\dfrac{3}{5}}\bigg)}^{2x+1}\bigg)}^{3}={\bigg(\dfrac{125}{27}\bigg)}^{3}}$

$\leadsto\tt{\bigg({\dfrac{3}{5}}\bigg)^{2x+1}={\bigg(\dfrac{5}{3}\bigg)}^{9}}$

$\leadsto\tt{\bigg({\dfrac{3}{5}}\bigg)^{2x+1}={\bigg(\dfrac{3}{5}\bigg)}^{(-9)}}$

So, 2x + 1 = -9

➜ 2x = - 10

➜ x = -5

$\rule{200}2$

Verification:

Putting the value of x in the question:

$\leadsto\tt{\bigg(\sqrt[3]{\dfrac{3}{5}}\bigg)^{2(-5)+1}=\dfrac{125}{27}}$

$\leadsto\tt{\bigg(\sqrt[3]{\dfrac{3}{5}}\bigg)^{(-9)}=\dfrac{125}{27}}$

$\leadsto\tt{\bigg({\dfrac{3}{5}}\bigg)^{(-9\times\frac{1}{3})}=\dfrac{125}{27}}$

$\leadsto\tt{\bigg({\frac{3}{5}}\bigg)^{(\dfrac{-9}{3})}=\dfrac{125}{27}}$

$\leadsto\tt{\bigg({\frac{3}{5}}\bigg)^{(-3)}=\dfrac{125}{27}}$

$\leadsto\tt{\bigg({\frac{5}{3}}\bigg)^{(3)}=\dfrac{125}{27}}$

Now, we observe that both left hand side and the right hand side are equal, hence verified.

$\rule{200}4$