# Using an Exponent to Undo Log

• Jun 28th 2013, 08:22 PM
Jason76
Using an Exponent to Undo Log
What is the next step in solving this (using this method)? But there is another way which involves using the log laws (which I know how to do).

$\log x - \dfrac{1}{3}\log8 = \log7$

$10^{\log(x)- \frac{1}{3}\log8} = 10^{\log7}$

I believe the next step involves dividing the bases of the exponents, since the exponents are subtracted.
• Jun 28th 2013, 09:57 PM
mpx86
Re: Using an Exponent to Undo Log
why r u doing this???

isnt log x=log7 +log2
use log a + log b= log ab
• Jun 28th 2013, 11:34 PM
Jason76
Re: Using an Exponent to Undo Log
An example of exponanted stuff:

$e^{\ln (2 + y^{2)}} = e^{\ln(4 + x^{2}) + C}$

$2 + y^{2} = (4 + x^{2}) e^{C}$ Since adding, then multiplying comes in.

But with the other example on this page, some subtracting was in there.
• Jun 29th 2013, 01:17 AM
ibdutt
Re: Using an Exponent to Undo Log
Do it with the laws of logarithms, you will get the answer straight unless you have a special purpose.
lox x - 1/3 log 8 = log 7
log x - log(8)^(1/3 ) = log 7
log x - log 2 = log 7
log ( x/2) = log 7 that will give x = 14
• Jun 29th 2013, 05:14 PM
Soroban
Re: Using an Exponent to Undo Log
Hello, Jason76!

Quote:

$\text{Solve for }x\!:\;\log x - \tfrac{1}{3}\log8 \:=\: \log7$

Simplify the logs before you exponentiate.

We have: . $\log x \:=\:\log 7 + \tfrac{1}{3}\log 8$

. . . . . . . . $\log x \;=\;\log 7 + \log\left(8^{\frac{1}{3}}\right)$

. . . . . . . . $\log x \;=\;\log 7 + \log 2$

. . . . . . . . $\log x \;=\;\log(7\cdot2)$

. . . . . . . . $\log x \;=\;\log 14$

Therefore: n . $x \;=\;14$