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Question Number 107153 by bemath last updated on 09/Aug/20

      @bemath@  (((14)/5))^(((28)/(√x))−5) = ((5/(14)))^((5/(√x))−160)

$$\:\:\:\:\:\:@{bemath}@ \\ $$$$\left(\frac{\mathrm{14}}{\mathrm{5}}\right)^{\frac{\mathrm{28}}{\sqrt{{x}}}−\mathrm{5}} =\:\left(\frac{\mathrm{5}}{\mathrm{14}}\right)^{\frac{\mathrm{5}}{\sqrt{{x}}}−\mathrm{160}} \\ $$

Answered by bobhans last updated on 09/Aug/20

       εbobhansε  (((14)/5))^(((28)/(√x))−5) = (((14)/5))^(160−(5/(√x)))   ⇔ ((28)/(√x))−5 = 160−(5/(√x)) ; [set (1/(√x)) = m , x>0]  ⇔28m−5 = 160−5m   ⇒ 33m = 165 ⇒m = 5 = (1/(√x)) . solution x = (1/(25))

$$\:\:\:\:\:\:\:\varepsilon\mathrm{bobhans}\varepsilon \\ $$$$\left(\frac{\mathrm{14}}{\mathrm{5}}\right)^{\frac{\mathrm{28}}{\sqrt{\mathrm{x}}}−\mathrm{5}} =\:\left(\frac{\mathrm{14}}{\mathrm{5}}\right)^{\mathrm{160}−\frac{\mathrm{5}}{\sqrt{\mathrm{x}}}} \\ $$$$\Leftrightarrow\:\frac{\mathrm{28}}{\sqrt{\mathrm{x}}}−\mathrm{5}\:=\:\mathrm{160}−\frac{\mathrm{5}}{\sqrt{\mathrm{x}}}\:;\:\left[\mathrm{set}\:\frac{\mathrm{1}}{\sqrt{\mathrm{x}}}\:=\:\mathrm{m}\:,\:\mathrm{x}>\mathrm{0}\right] \\ $$$$\Leftrightarrow\mathrm{28m}−\mathrm{5}\:=\:\mathrm{160}−\mathrm{5m} \\ $$$$\:\Rightarrow\:\mathrm{33m}\:=\:\mathrm{165}\:\Rightarrow\mathrm{m}\:=\:\mathrm{5}\:=\:\frac{\mathrm{1}}{\sqrt{\mathrm{x}}}\:.\:\mathrm{solution}\:\mathrm{x}\:=\:\frac{\mathrm{1}}{\mathrm{25}} \\ $$

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