Question

find the value of x if 4^x + 6^x = 9^x​

Answer 1

Question :

$\bullet\ \quad \sf{\red{4^x+6^x=9^x}}$

Solution :

$\sf 4^x+6^x=9^x$

Divide both sides of the eq. with 4ˣ .

$\longrightarrow \sf \dfrac{(4^x+6^x)}{4^x}=\dfrac{9^x}{4^x}$

$\longrightarrow \sf \dfrac{4^x}{4^x}+\dfrac{6^x}{4^x}=\dfrac{9^x}{4^x}$

$\longrightarrow \sf 1+\left(\dfrac{6}{4}\right)^x=\left(\dfrac{9}{4}\right)^x$

$\longrightarrow \sf 1+\left(\dfrac{3}{2}\right)^x=\left(\left(\dfrac{3}{2}\right)^2\right)^x$

$\longrightarrow \sf 1+\left(\dfrac{3}{2}\right)^x=\left(\dfrac{3}{2}\right)^{2x}$

$\longrightarrow \sf \left(\left(\dfrac{3}{2}\right)^{x}\right)^2-\left(\dfrac{3}{2}\right)^x-1=0$

Let  $\sf \left( \dfrac{3}{2}\right)^x=p$ ,

$\longrightarrow\ \sf p^2-p-1=0$

It's a quadratic eq. of p in the form ax² + bx + c ,

where , a = 1 , b = - 1 , c = - 1

$\longrightarrow\ \sf p= \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$

$\longrightarrow\ \sf p= \dfrac{-(-1) \pm \sqrt{(-1)^2-4(1)(-1)}}{2(1)}$

$\longrightarrow\ \sf p= \dfrac{1 \pm \sqrt{1+4}}{2}$

$\longrightarrow\ \sf p= \dfrac{1 \pm \sqrt{5}}{2}$

For positive solution ,

$\longrightarrow\ \sf p= \dfrac{1 + \sqrt{5}}{2}$

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

Applying logarithm ,

$\longrightarrow\ \sf \ln \left( \dfrac{3}{2}\right)^x= \ln \left(\dfrac{1 + \sqrt{5}}{2} \right)$

$\bullet\ \quad \sf \orange{\ln a^b=b \ln a}$

$\longrightarrow\ \sf x\ \ln \left( \dfrac{3}{2}\right)= \ln \left(\dfrac{1 + \sqrt{5}}{2} \right)$

$\bullet\ \quad \sf \green{\ln \dfrac{a}{b}=\ln a-\ln b}$

$\longrightarrow\ \sf \pink{x= \dfrac{\ln (1 + \sqrt{5}) - \ln 2}{ \ln 3 - \ln 2}}$

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Answer 2

Question :

$\bullet\ \quad \sf{\red{4^x+6^x=9^x}}$

Solution :

$\sf 4^x+6^x=9^x$

Divide both sides of the eq. with 4ˣ .

$\longrightarrow \sf \dfrac{(4^x+6^x)}{4^x}=\dfrac{9^x}{4^x}$

$\longrightarrow \sf \dfrac{4^x}{4^x}+\dfrac{6^x}{4^x}=\dfrac{9^x}{4^x}$

$\longrightarrow \sf 1+\left(\dfrac{6}{4}\right)^x=\left(\dfrac{9}{4}\right)^x$

$\longrightarrow \sf 1+\left(\dfrac{3}{2}\right)^x=\left(\left(\dfrac{3}{2}\right)^2\right)^x$

$\longrightarrow \sf 1+\left(\dfrac{3}{2}\right)^x=\left(\dfrac{3}{2}\right)^{2x}$

$\longrightarrow \sf \left(\left(\dfrac{3}{2}\right)^{x}\right)^2-\left(\dfrac{3}{2}\right)^x-1=0$

Let  $\sf \left( \dfrac{3}{2}\right)^x=p$ ,

$\longrightarrow\ \sf p^2-p-1=0$

It's a quadratic eq. of p in the form ax² + bx + c ,

where , a = 1 , b = - 1 , c = - 1

$\longrightarrow\ \sf p= \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$

$\longrightarrow\ \sf p= \dfrac{-(-1) \pm \sqrt{(-1)^2-4(1)(-1)}}{2(1)}$

$\longrightarrow\ \sf p= \dfrac{1 \pm \sqrt{1+4}}{2}$

$\longrightarrow\ \sf p= \dfrac{1 \pm \sqrt{5}}{2}$

For positive solution ,

$\longrightarrow\ \sf p= \dfrac{1 + \sqrt{5}}{2}$

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

Applying logarithm ,

$\longrightarrow\ \sf \ln \left( \dfrac{3}{2}\right)^x= \ln \left(\dfrac{1 + \sqrt{5}}{2} \right)$

$\bullet\ \quad \sf \orange{\ln a^b=b \ln a}$

$\longrightarrow\ \sf x\ \ln \left( \dfrac{3}{2}\right)= \ln \left(\dfrac{1 + \sqrt{5}}{2} \right)$

$\bullet\ \quad \sf \green{\ln \dfrac{a}{b}=\ln a-\ln b}$

$\longrightarrow\ \sf \pink{x= \dfrac{\ln (1 + \sqrt{5}) - \ln 2}{ \ln 3 - \ln 2}}$

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