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

[phosgene]

Homework Statement

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

$log_a(b)$

Homework Equations

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

The Attempt at a Solution

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

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

 

[Sourabh N]

A useful relation is log_a b = log_eb / log_ea.
How is log_eb related to e^b?

 

[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:

log_a(b)=lnb/lna

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

I can re-arrange so that I get exp(log_a(b))=(exp(lnb))1/lna)=b^1/lna, then I'm unsure where to go as I don't know what to do with the 1/lna.

 

[Sourabh N]

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:

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

log_a(b)=lnb/lna

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

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

I can re-arrange so that I get exp(log_a(b))=(exp(lnb))1/lna)=b^1/lna, then I'm unsure where to go as I don't know what to do with the 1/lna.

You don't need to do this.

 

[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 :)

 

[uart]

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 :)

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

 

[HallsofIvy]

Homework Statement

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

log_a(b)

Homework Equations

ln(b)=log_a(b)/log_a(e)

The Attempt at a Solution

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

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

(Do NOT put the HTML "sub" tags inside "tex" tags, either do not use "tex" or use underscores, "_", in "tex".)
For any x, e^ln(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 $y= ln_a(b)= ln(b)/ln(a)$.