# Thread: Logarithms with an x,y, and z values?

1. ## Logarithms 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?

2. you should know that

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

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

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

apply these two formula you will get the answer

3. Originally Posted by Amer
you should know that

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

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

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

apply these two formula you will get the answer
If only it was that easy for me. Using that I can get it down to
log7(x^(1/3)*y^(13)) -5log5^z

What is throwing me off is the different bases. 5 and 7. And then the different variables x,y, and a third variable z.

4. Originally Posted by bilbobaggins
Take this one for example
1/3log7^x +13log7^y - 5log5^z

How would I reduce this to a single logarithm?
see

$1/3 log(7^x)$

$log(7^{1/3x})$

$13log(7^y)=??$

5. Originally Posted by Amer
see

$1/3 log(7^x)$

$log(7^{1/3x})$

$13log(7^y)=??$
Oh, I just realized that before I refreshed and saw your response. Thanks.

6. the number the is multiplies with the Logarithm is a power for what is inside it

see this

$a(log(b))=log(a^b)$

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

so

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

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

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

now

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

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

7. Originally Posted by Amer
the number the is multiplies with the Logarithm is a power for what is inside it

see this

$a(log(b))=log(a^b)$

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

so

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

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

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

now

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

$log 7^{\frac{x}{3}}+log 7^{13y}-log 5^{5z}=log\dfrac{\left(7^{\frac{x}{3}}\right)\left (7^{13y}\right)}{5^{5z}}$
So you can just take the base out of the log and put it in a fraction? Thanks, I didn't know that.

8. Originally Posted by bilbobaggins
So you can just take the base out of the log and put it in a fraction?
No.

Are you saying that your 7s were actually the bases and the stuff you presented as exponents were really arguments?

9. Originally Posted by bilbobaggins
Take this one for example
1/3log7^x +13log7^y - 5log5^z

How would I reduce this to a single logarithm?
If those numbers after "log" are the base (ie: $\frac{1}{3}log_7(x) + 13log_7(y) - 5log_5(z)$ then you can combine the first two terms using the laws written by Amer

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

Use the change of base rule on this new term and the second term to make them the same base

10. Originally Posted by e^(i*pi)
If those numbers after "log" are the base (ie: $\frac{1}{3}log_7(x) + 13log_7(y) - 5log_5(z)$ then you can combine the first two terms using the laws written by Amer

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

Use the change of base rule on this new term and the second term to make them the same base
So (log(x^1/3*y^13)/log7) - (log(z^5)/log5)?

11. Originally Posted by stapel
No.

Are you saying that your 7s were actually the bases and the stuff you presented as exponents were really arguments?
Yeah pretty much, sorry. I don't know exactly how to put logarithms into a computer.

12. change bases in logarithms

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

so you can write

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

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

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

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

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

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

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