Solve for b,
y=b^x

Question

Solve for b,
y=b^x

anonymous · Answer

$$y = b^{x} \Leftrightarrow y^{\frac{1}{x}} = b$$

anonymous · Answer

thats taking the x-th root of y

anonymous · Answer

$$b=\sqrt[x]{y}$$

anonymous · Answer

wow, that was a lot easier than what I was trying to do. Thanks

anonymous · Answer

b= x-th root of y

anonymous · Answer

If you need the steps between.
$$\ln y = \ln b^x$$
$$\ln y = x \ln b$$
$$1/x \ln y = \ln b$$
$$\ln y^{1/x} = \ln b$$
$$e^{\ln y^{1/x}} = e^{\ln b}$$
$$y^{1/x} = b$$

anonymous · Answer

Thats making it way too complicated >.< no need for logarithmic functions.  Just like you take the square root of something squared, or you take the cube root of something cube, you would take the x-th root of something to the x power.

anonymous · Answer

You're right, but honestly whenever I see something raised to the x I automatically think of logarithms.

anonymous · Answer

touche, at first i thought he said solve for x so i was all over the ln lol

anonymous · Answer

I went for the logs originally, then went wrong trying to take the derivative, thanks for the help