Question

Solve the given equation for x.
Pls help me with this question​
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Answer 1

Answer:

$\boxed{\huge{\pink{x = e^{2}}}}$

Step-by-step explanation:

$\bold{Given}\longrightarrow \\ {5}^{logx} - {4}^{logx - 1} = {4}^{logx} + {5}^{logx - 1}$

$\bold{To \: find}\longrightarrow \\ value \: of \: x$

$\bold{Concept \: used}\longrightarrow \\ \\ 1) {a}^{m} {b}^{m} = {(ab)}^{m} \\ 2) log_{m}(x) = n \\ = > x = {m}^{n}$

$\bold{Solution}\longrightarrow \\ {5}^{logx} - {4}^{logx - 1} = {4}^{logx} + {5}^{logx - 1}$

$= > {5}^{logx} - {5}^{lox - 1} = {4}^{logx} + {4}^{logx - 1} \\ = > {5}^{logx} - {5}^{logx} \: {5}^{ - 1} = {4}^{logx} + {4}^{logx} \: {4}^{ - 1}$

$= > {5}^{logx} (1 - {5}^{ - 1} ) = {4}^{logx} (1 + {4}^{ - 1} )$

$= > {5}^{logx} (1 - \dfrac{1}{5} ) = {4}^{logx} (1 + \dfrac{1}{4} )$

$= > {5}^{logx} ( \dfrac{4}{5} ) = {4}^{logx} ( \dfrac{5}{4} )$

$= > \dfrac{ {5}^{logx} }{ {4}^{lox} } = \dfrac{25}{16}$

$= > {( \dfrac{5}{4}) }^{logx} = {( \dfrac{5}{4} )}^{2}$

$comparing \: exponent \: we \: get$

$= > logx \: = \: 2$

$= > log_{e}(x) = 2$

$= > x \: = {e}^{2}$