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:

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