Question

${9}^{x} + {15}^{x} = {25}^{x}$
find the value of x​

Answer 1

$x=\dfrac{\log2-\log(\sqrt{5}-1)}{\log5-\log3}$

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$\Large\textrm{\underline{\underline{Question}}}$

$9^{x}+15^{x}=25^{x}$

Solve for real $x$.

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$\Large\textrm{\underline{\underline{Main Topics}}}$

We need to understand the laws of exponent, quadratic formula, and logarithm to solve this question.

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$\large\textrm{Laws of Exponents}$

$\boxed{a^{n}\cdot b^{n}=(ab)^{n}}$

$\boxed{(a^{m})^{n}=a^{mn}}$

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$\large\textrm{Quadraic Formula}$

$\boxed{ax^{2}+bx+c=0\ (a\neq0)\iff x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}}$

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$\large\textrm{Logarithm}$

$\boxed{\log_{a^{m}}b^{n}=\dfrac{n}{m}\log_{a}b}$

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$\Large\textrm{\underline{\underline{Solution}}}$

Assumption:

• $A=3^{x}$
• $B=5^{x}$

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Now

$9^{x}+15^{x}=25^{x}$

$\Longrightarrow(3^{2})^{x}+(3\cdot5)^{x}=(5^{2})^{x}$

$\Longrightarrow 3^{2x}+3^{x}\cdot 5^{x}=5^{2x}$

[By assumption, we get the following.]

$\Longrightarrow A^{2}+AB=B^{2}$

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Now

$A^{2}+AB=B^{2}$

[Divide both sides by $B^{2}$.]

$\Longrightarrow \left(\dfrac{A}{B}\right)^{2}+\dfrac{A}{B}=1$

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Assumption:-

• $C=\dfrac{A}{B}$

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Now

$\left(\dfrac{A}{B}\right)^{2}+\dfrac{A}{B}=1$

$\Longrightarrow C^{2}+C=1$

$\Longrightarrow C^{2}+C-1=0$

[Now, this is a quadratic equation in $C$.]

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The quadratic formula is given by

$\boxed{ax^{2}+bx+c=0\ (a\neq0)\iff x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}}$

$\Longrightarrow C=\dfrac{-1\pm\sqrt{5}}{2}$

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Remark:-

• $A=3^{x}$
• $B=5^{x}$
• $C=\dfrac{A}{B}$

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Now

$C=\left(\dfrac{3}{5}\right)^{x},\ \boxed{C > 0}$

[In disregarding non-real solutions.]

$\Longrightarrow \left(\dfrac{3}{5}\right)^{x}=\dfrac{-1+\sqrt{5}}{2}$

$\Longrightarrow x\log\dfrac{3}{5}=\log\dfrac{-1+\sqrt{5}}{2}$

$\Longrightarrow x=\dfrac{\log(\sqrt{5}-1)-\log2}{\log3-\log5}$

$\Longrightarrow\boxed{x=\dfrac{\log2-\log(\sqrt{5}-1)}{\log5-\log3}}$

[This is the required answer.]