Exponents with Variables

Rae Kirk

New member
Joined
Jan 14, 2020
Messages
2
Can’t find info on how to solve exponents with variables when bases are different.

(5^x)(3^(1-x))=12
 

MarkFL

Super Moderator
Staff member
Joined
Nov 24, 2012
Messages
2,412
Hello, and welcome to FMH! :)

I'd write the equation as:

\(\displaystyle \left(\frac{5}{3}\right)^x=4\)

Now, convert from exponential to logarithmic form...what do you get?
 

Rae Kirk

New member
Joined
Jan 14, 2020
Messages
2
That’s it!! I was making it too complicated. Thank you so much. What a great support group.
 

JeffM

Elite Member
Joined
Sep 14, 2012
Messages
4,238
Mark's approach is elegant, but you can just start by immediately going to logs.

\(\displaystyle (5^x)(3^{(1-x)}) = 12 \implies log\{(5^x)(3^{(1-x)})\} = log(12).\)

It is a bit longer, but it is totally mechanical.
 

Jomo

Elite Member
Joined
Dec 30, 2014
Messages
5,108
Hello, and welcome to FMH! :)

I'd write the equation as:

\(\displaystyle \left(\frac{5}{3}\right)^x=4\)

Now, convert from exponential to logarithmic form...what do you get?
Even low level math can look beautiful. Nicely done!
 
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