How do you solve a^x = 10^(2x+1)ax=102x+1?

1 Answer
Apr 5, 2016

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

Explanation:

11. Assuming you are trying to solve for xx, start by taking the logarithm of both sides.

a^x=10^(2x+1)ax=102x+1

log(a^x)=log(10^(2x+1))log(ax)=log(102x+1)

22. Using the logarithmic property, log_color(purple)b(color(red)m^color(blue)n)=color(blue)n*log_color(purple)b(color(red)m)logb(mn)=nlogb(m), simplify the equation.

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

33. Expand the brackets.

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

44. Move all terms with xx to one side of the equation with the terms with no xx to the other side.

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

55. Factor out xx.

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

66. Isolate for xx.

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