Logarithm Problem URGENT!!!

**aussiekid90** · January 30th 2006, 11:31 AM

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.

**ThePerfectHacker** · January 30th 2006, 12:42 PM

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.

$\log 5^x+1=\log 3^x$ , by the exponent rule for logarithms you may "bring down" the $x$ thus,
$x\log 5+1=x\log 3$
Then,
$x\log 5-x\log 3=-1$
Thus,
$x(\log 5-\log 3)=-1$
Thus,
$x=-\frac{1}{\log 5-\log 3}=\frac{1}{\log 3-\log 5}$
Q.E.D.

**aussiekid90** · January 30th 2006, 04:15 PM

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

**ThePerfectHacker** · January 30th 2006, 04:58 PM

Originally Posted by aussiekid90

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

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

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