# Finding the value of an exponential given that e^a=10 and e^b=4

1. Oct 8, 2012

### phosgene

1. The problem statement, all variables and given/known data

Suppose that e^a=10 and e^b=4. Find the value of

$loga(b)$

2. Relevant equations

$ln(b)=loga(b)/loga(e)$

3. The attempt at a solution

So far I've only gotten to the above equation rearranged to:

$ln(b) loga(e)=loga(b)/$. I'm unsure as to where I'm supposed to go from here...

2. Oct 8, 2012

### Sourabh N

A useful relation is loga b = logeb / logea.
How is logeb related to eb?

3. Oct 8, 2012

### phosgene

Hmm, you can get e^b from it by going exp(exp(ln(b))), but I'm kinda stuck. I get up to this stage:

loga(b)=lnb/lna

exp(loga(b))=exp(lnb/lna)

I can re-arrange so that I get exp(loga(b))=(exp(lnb))1/lna)=b^1/lna, then I'm unsure where to go as I dont know what to do with the 1/lna.

4. Oct 8, 2012

### Sourabh N

EDIT: That doesn't seem useful for the problem here. What you should use is b = ln eb.

Find ln a just the way you found ln b and substitute it here.

You don't need to do this.

5. Oct 8, 2012

### phosgene

Wait, I just realized that all I had to do was solve for a and b in the original equations and plug it in. Doh! Thanks for the help though :)

6. Oct 8, 2012

### uart

And if you want to write it in terms of 4 and 10 then it's $\frac{\ln \ln 4}{ \ln \ln 10}$

7. Oct 8, 2012

### HallsofIvy

Staff Emeritus
(Do NOT put the HTML "sub" tags inside "tex" tags, either do not use "tex" or use underscores, "_", in "tex".)
For any x, eln(x)= x so that $a^x= e^{ln(a^x)}= e^{x ln(a)}$. Similarly, if $y= log_a(b)$, then $b= a^y$ so that $b= e^{ln a^y}= e^{y ln a}$. Then $ln(b)= y ln(a)$ so that $y= ln_a(b)= ln(b)/ln(a)$.