Answer to Question #171294 in Algebra for gibson

Question

solve for x in the following equations

1. 4^x-2 = 3^x+4
2. 9^x-3^x-8=0

Answer 1

1) Let's take logarithm of both sides of equation:

$log(4^{x-2})=log(3^{x+4}),$ $(x-2)\cdot log(4)=(x+4)\cdot log(3),$ $xlog(4)-2log(4)=xlog(3)+4log(3),$ $xlog(4)-xlog(3)=4log(3)+2log(4),$ $x(log(4)-log(3))=4log(3)+2log(4),$ $x(log(\dfrac{4}{3}))=log(81)+log(16),$ $x(log(\dfrac{4}{3}))=log(1296),$ $x=\dfrac{log(1296)}{log(\dfrac{4}{3})}.$

Answer:

$x=\dfrac{log(1296)}{log(\dfrac{4}{3})}.$

2)

$9^x-3^x-8=0,$ $3^{2x}-3^x-8=0.$

Let $3^x=y$, then we get the following quadratic equation:

$y^2-y-8=0.$

This quadratic equation has two roots:

$y_1=\dfrac{-b+\sqrt{b^2-4ac}}{2a}=\dfrac{1+\sqrt{33}}{2},$ $y_2=\dfrac{-b-\sqrt{b^2-4ac}}{2a}=\dfrac{1-\sqrt{33}}{2}.$

Let's go back, substitute $y$ and solve for $x$:

$3^x=\dfrac{1+\sqrt{33}}{2},$ $log(3^x)=log(\dfrac{1+\sqrt{33}}{2}),$ $xlog(3)=log(\dfrac{1+\sqrt{33}}{2}),$ $x=\dfrac{log(\dfrac{1+\sqrt{33}}{2})}{log(3)}.$

There is no solution for case when we substitute $y_2=\dfrac{1-\sqrt{33}}{2}.$

Answer:

$x=\dfrac{log(\dfrac{1+\sqrt{33}}{2})}{log(3)}.$