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.

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- Jan 30th 2006, 11:31 AMaussiekid90Logarithm 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 PMThePerfectHackerQuote:

Originally Posted by**aussiekid90**

$\displaystyle x\log 5+1=x\log 3$

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 PMaussiekid90
Sorry, typo, the problem would actually be 5^(x+1)=3^x

- Jan 30th 2006, 04:58 PMThePerfectHackerQuote:

Originally Posted by**aussiekid90**

$\displaystyle \log 5^{x+1}=\log 3^x$

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.