Take this one for example

1/3log7^x +13log7^y - 5log5^z

How would I reduce this to a single logarithm?

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- Jun 5th 2009, 12:32 PMbilbobagginsLogarithms with an x,y, and z values?
Take this one for example

1/3log7^x +13log7^y - 5log5^z

How would I reduce this to a single logarithm? - Jun 5th 2009, 12:58 PMAmer
you should know that

$\displaystyle log(a)^b=blog(a)$

$\displaystyle log(a)-log(b)=log(\frac{a}{b})$

$\displaystyle log(a)+log(b)=log(ab)$

apply these two formula you will get the answer - Jun 5th 2009, 01:04 PMbilbobaggins
- Jun 5th 2009, 01:07 PMAmer
- Jun 5th 2009, 01:17 PMbilbobaggins
- Jun 5th 2009, 01:31 PMAmer
the number the is multiplies with the

**Logarithm is a power for what is inside it**$\displaystyle a(log(b))=log(a^b)$

see this

$\displaystyle a(log(b)^c)=log(b^{ac})$

so

$\displaystyle \frac{1}{3} log 7^x = log 7^{\frac{x}{3}}$

$\displaystyle 13 log 7^y = log 7^{13y}$

$\displaystyle -5log 5^z=-log 5^{5z}$

now

$\displaystyle log 7^{\frac{x}{3}}+log 7^{13y}-log 5^{5z}=log\left(7^{\frac{x}{3}}\right)\left(7^{13y }\right)-log 5^{5z}$

$\displaystyle log 7^{\frac{x}{3}}+log 7^{13y}-log 5^{5z}=log\dfrac{\left(7^{\frac{x}{3}}\right)\left (7^{13y}\right)}{5^{5z}}$ - Jun 5th 2009, 07:56 PMbilbobaggins
- Jun 6th 2009, 05:04 AMstapel
- Jun 6th 2009, 05:48 AMe^(i*pi)
If those numbers after "log" are the base (ie: $\displaystyle \frac{1}{3}log_7(x) + 13log_7(y) - 5log_5(z)$ then you can combine the first two terms using the laws written by Amer

$\displaystyle log_7(x^{\frac{1}{3}}y^{13})$

Use the change of base rule on this new term and the second term to make them the same base - Jun 6th 2009, 08:22 AMbilbobaggins
- Jun 6th 2009, 08:23 AMbilbobaggins
- Jun 6th 2009, 09:51 AMAmer
change bases in logarithms

$\displaystyle log_{a}b=\frac{log_c b}{log_c a}$

so you can write

$\displaystyle -5log_{5}z=-log_5 z^5=-\frac{log_{7}(z^5)}{log_{7}(5)}$

to change it base so you can apply what I mention before

http://www.mathhelpforum.com/math-he...7b29a014-1.gif

$\displaystyle log_{7}(x^{\frac{1}{3}}y^{13})-\frac{log_{7}(z^5)}{\log_{7}(5)}$

$\displaystyle {\log_{7}(5)}$ consider as constant right

$\displaystyle log_{7}(x^{\frac{1}{3}}y^{13})-log_{7}(z^5)(\log_{7}(5))^{-1}$

$\displaystyle log_{7}(x^{\frac{1}{3}}y^{13})-log_{7}(z^{5(\log_{7}(5))^{-1}})$

$\displaystyle log_{7}\left(\frac{(x^{\frac{1}{3}}y^{13})}{(z^{5( \log_{7}(5))^{-1}})}\right)$