# Solve without using log

• Aug 28th 2010, 07:47 PM
Punch
Solve without using log
$\displaystyle solve \frac{3^n}{49}=1029^n$ without using logarithms
• Aug 28th 2010, 08:03 PM
Prove It
$\displaystyle \frac{3^n}{49} = 1029^n$

$\displaystyle \frac{3^n}{1029^n} = 49$

$\displaystyle \left(\frac{3}{1029}\right)^n = 49$

$\displaystyle \left(\frac{1}{343}\right)^n = 49$

$\displaystyle \frac{1^n}{343^n} = 7^2$

$\displaystyle \frac{1}{(7^{3})^n} = 7^2$

$\displaystyle \frac{1}{7^{3n}} = 7^2$

$\displaystyle 7^{-3n} = 7^2$

$\displaystyle -3n = 2$

$\displaystyle n = -\frac{2}{3}$.
• Aug 28th 2010, 08:04 PM
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Quote:

Originally Posted by Punch
$\displaystyle solve \frac{3^n}{49}=1029^n$ without using logarithms

Hint: 1029 = 3*7*7*7. Rewrite RHS in a way that allows you to cancel something out.

Edit: Too slow, so might as well show all the steps I had in mind.

$\displaystyle \frac{3^n}{49}=1029^n$

$\displaystyle 3^n7^{-2}=3^n7^{3n}$

$\displaystyle 7^{-2}=7^{3n}$

$\displaystyle -2=3n$

$\displaystyle n=-\frac{2}{3}$