Calculus 1 Solving logarithmic equation

[destinc]

Solve for b,
f(x)=b^x

Here is where I am so far
lny=lnb^x
lny=xlnb^x
y'/y= lnb
e^y'/y= e^lnb
b= e^y'/y

My question is, can I simplify the exponent y'/y any further, or is my answer good here?
Thank you

 

[Dschumanji]

You say to solve for b in the function [itex]f\left(x\right) = b^{x}[/itex]. From the answer provided you solve for b in terms of [itex]f\left(x\right)[/itex] and [itex]f'\left(x\right)[/itex]. Does the original problem ask for you to solve for b in terms of these functions? It is possible to solve for b in terms of only [itex]x[/itex] and [itex]f\left(x\right)[/itex].

 

[destinc]

It does not require taking the derivative, that was simply the first idea I had. If there is an alternate way, please show me

 

[Dschumanji]

Look at the second line of the work that you have already provided. It should actually be [itex]ln\left(y\right) = xln\left(b\right)[/itex]. If x is not equal to zero, then you can divide both sides by x. The rest of the simplification is left to you.

 

[SammyS]

Solve for b,
f(x)=b^x

Here is where I am so far
lny=lnb^x
lny=xlnb^x This line should be lny=xlnb. It shouldn't have b^x.
y'/y= lnb
e^y'/y= e^lnb
b= e^y'/y

My question is, can I simplify the exponent y'/y any further, or is my answer good here?
Thank you

Rather than using logs, take both sides of the equation, y=b^x, to the 1/x power.

 

[destinc]

thanks, I made the problem so much harder than I needed to. The redirect helped.