Answer to Question #155275 in Algebra for Tony

Question

Solve the equation:
4^(x) + 6^(x) = 9^(x)

Answer 1

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

Will divide by $4^x$

$1+\frac{6^x}{4^x}=\frac{9^x}{4^x}\\1+(\frac{3}{2})^x=(\frac{3}{2})^{2x}$

Put $(\frac{3}{2})^x=t$

Then

$t^2-t-1=0$

$t=\frac{1 \mp \sqrt{1^2-4\times1\times(-1)}}{2\times1}\\t=\frac{1\mp \sqrt{5}}{2}\\$

t should have only positive value because 3/2 is positive value and power of positive number always give positive value.

Thus

$t=\frac{1+\sqrt{5}}{2}$

$\frac{1+\sqrt{5}}{2}=(\frac{3}{2})^x \\and\\ln(\frac{1+\sqrt{5}}{2})=xln(\frac{3}{2})\\x=1.23$