Question

Find $x$ if $4^x+6^x=9^x$

• A. $\frac{ln(\sqrt{3}-1)+ln2}{ln2-ln3}$
• B. $\frac{ln(\sqrt{5}-1)+ln2}{ln2-ln3}$
• C. $\frac{ln(\sqrt{5}-1)-ln2}{ln2-ln3}$
• D. $\frac{ln(\sqrt{5}-1)+ln2}{ln3-ln2}$

Answer 1

The correct option is A $\frac{ln(\sqrt{3}-1)+ln2}{ln2-ln3}$

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

divide by $4^x$ to form ${\left(\frac{3}{2}\right)}^x$

$\frac{6^x}{4^x}+1=\frac{9^x}{4^x}$

${\left(\frac{3}{2}\right)}^x+1={\left(\frac{3}{2}\right)}^{2x}$

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

So ${\left(\frac{3}{2}\right)}^x=\frac{1\pm \sqrt{1-4.1(-1)}}{2}=\frac{1\pm ,\sqrt{5}}{2}$

For positive solution

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

Applying logarithm

$xln\left(\frac{3}{2}\right)=ln\left(\frac{1+\sqrt{5}}{2}\right)$

$ln\frac{a}{b}=lna-lnb$

$x=\frac{ln(1+\sqrt{5})-ln2}{ln3-ln2}$.