Question

How do you solve `a^x = 10^(2x+1)`?

Answer 1

`x=(-log(10))/(2log(10)-log(a))`

Explanation:

`1`. Assuming you are trying to solve for `x`, start by taking the logarithm of both sides.

`a^x=10^(2x+1)`

`log(a^x)=log(10^(2x+1))`

`2`. Using the logarithmic property, `log_color(purple)b(color(red)m^color(blue)n)=color(blue)n*log_color(purple)b(color(red)m)`, simplify the equation.

`xlog(a)=(2x+1)log(10)`

`3`. Expand the brackets.

`xlog(a)=2xlog(10)+log(10)`

`4`. Move all terms with `x` to one side of the equation with the terms with no `x` to the other side.

`2xlog(10)-xlog(a)=-log(10)`

`5`. Factor out `x`.

`x(2log(10)-log(a))=-log(10)`

`6`. Isolate for `x`.

`color(green)(|bar(ul(color(white)(a/a)x=(-log(10))/(2log(10)-log(a))color(white)(a/a)|)))`