Question

The value of $x$, which satisfies $4^x+6^x=9^x$ is

• A. $\frac{\ln (\sqrt{5}-1)-\ln 2}{\ln 3-\ln 2}$
• B. $\frac{\ln (\sqrt{5}-1)+\ln 2}{\ln 2-\ln 3}$
• C. $\frac{\ln (\sqrt{5}-1)-\ln 2}{\ln 2-\ln 3}$
• D. $\frac{\ln (\sqrt{5}-1)+\ln 2}{\ln 3-\ln 2}$

Answer 1

The correct option is C $\frac{\ln (\sqrt{5}-1)-\ln 2}{\ln 2-\ln 3}$

$4^x+6^x=9^x$
$\Rightarrow {\left(\frac{4}{9}\right)}^x+{\left(\frac{2}{3}\right)}^x=1$
Substitute ${\left(\frac{2}{3}\right)}^x=y$
$\Rightarrow y^2+y-1=0$
$\Rightarrow y=\frac{-1\pm \sqrt{5}}{2}$
$\Rightarrow {\left(\frac{2}{3}\right)}^x=\frac{\sqrt{5}-1}{2}(\because$ Exponential value cannot be negative.)
$\Rightarrow x\ln \left(\frac{2}{3}\right)=\ln \frac{\sqrt{5}-1}{2}$
$\Rightarrow x=\frac{\ln (\sqrt{5}-1)-\ln 2}{\ln 2-\ln 3}$
Hence, option 'C' is correct.