# Calculus 1 Solving logarithmic equation

1. Jul 3, 2011

### 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

2. Jul 3, 2011

### Dschumanji

You say to solve for b in the function $f\left(x\right) = b^{x}$. From the answer provided you solve for b in terms of $f\left(x\right)$ and $f'\left(x\right)$. 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 $x$ and $f\left(x\right)$.

3. Jul 3, 2011

### 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

4. Jul 3, 2011

### Dschumanji

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

5. Jul 3, 2011

### SammyS

Staff Emeritus
Rather than using logs, take both sides of the equation, y=bx, to the 1/x power.

6. Jul 4, 2011

### destinc

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

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