# Logarithm Problem URGENT!!!

• Jan 30th 2006, 11:31 AM
aussiekid90
Logarithm Problem URGENT!!!
I have had a problem with logarithm problems and one in particular; the problem states: log 5^x+1= log 3^x. Can you please help me with this problem and explain the steps as well? Thank you.
• Jan 30th 2006, 12:42 PM
ThePerfectHacker
Quote:

Originally Posted by aussiekid90
I have had a problem with logarithm problems and one in particular; the problem states: log 5^x+1= log 3^x. Can you please help me with this problem and explain the steps as well? Thank you.

$\displaystyle \log 5^x+1=\log 3^x$, by the exponent rule for logarithms you may "bring down" the $\displaystyle x$ thus,
$\displaystyle x\log 5+1=x\log 3$
Then,
$\displaystyle x\log 5-x\log 3=-1$
Thus,
$\displaystyle x(\log 5-\log 3)=-1$
Thus,
$\displaystyle x=-\frac{1}{\log 5-\log 3}=\frac{1}{\log 3-\log 5}$
Q.E.D.
• Jan 30th 2006, 04:15 PM
aussiekid90
Sorry, typo, the problem would actually be 5^(x+1)=3^x
• Jan 30th 2006, 04:58 PM
ThePerfectHacker
Quote:

Originally Posted by aussiekid90
Sorry, typo, the problem would actually be 5^(x+1)=3^x

Take the logarithm of both sides,
$\displaystyle \log 5^{x+1}=\log 3^x$
Now use the exponent rule and "bring down" exponents,
$\displaystyle (x+1)\log 5=x\log 3$
Open parantheses,
$\displaystyle x\log 5+\log 5=x\log 3$
Rearrange,
$\displaystyle x\log 5-x\log 3=-\log 5$
Factor,
$\displaystyle x(\log 5-\log 3)=-\log 5$
Thus,
$\displaystyle x=-\frac{\log 5}{\log 5-\log 3}\approx -3.15$
Q.E.D.