# Math Help - logarithms help

1. ## logarithms help

Solve the simultaneous equations $4^x=5^y$ and $2(4^x)=7^y$.

I have tried substituting $5^y$ into $2(4^x)=7^y$ and also changing $4^x=5^y$ into $y=log_{5}4^x$. I seem to be going on and on without really going anywhere with my working.
Could really do with some help. Thanks

2. Originally Posted by greghunter
Solve the simultaneous equations $4^x=5^y$ and $2(4^x)=7^y$.

I have tried substituting $5^y$ into $2(4^x)=7^y$ and also changing $4^x=5^y$ into $y=log_{5}4^x$. I seem to be going on and on without really going anywhere with my working.
Could really do with some help. Thanks
You have $2(5^y)=7^y$, so taking logs (to any base) you have:

$\log(2)+y\log(5)=y\log(7)$

hence:

$y= ...$

CB

3. Originally Posted by greghunter
Solve the simultaneous equations $4^x=5^y$ and $2(4^x)=7^y$.

I have tried substituting $5^y$ into $2(4^x)=7^y$ and also changing $4^x=5^y$ into $y=log_{5}4^x$. I seem to be going on and on without really going anywhere with my working.
Could really do with some help. Thanks
$2\left(4^x\right) = 7^y$ and $4^x = 5^y$.

Therefore

$2\left(5^y\right) = 7^y$

$2 = \frac{7^y}{5^y}$

$2 = \left(\frac{7}{5}\right)^y$

$\ln{2} = \ln{\left(\frac{7}{5}\right)^y}$

$\ln{2} = y\ln{\frac{7}{5}}$

$\ln{2} = y(\ln{7} - \ln{5})$

$y = \frac{\ln{2}}{\ln{7} - \ln{5}}$.

Can you solve for $x$?

4. Thanks, I didn't realise you could stick a log infront but it does make sense

5. Originally Posted by greghunter
Solve the simultaneous equations $4^x=5^y$ and $2(4^x)=7^y$.

I have tried substituting $5^y$ into $2(4^x)=7^y$ and also changing $4^x=5^y$ into $y=log_{5}4^x$. I seem to be going on and on without really going anywhere with my working.
Could really do with some help. Thanks

Divide the second equation by the first giving 2 = (7^y)/(5^y)
then 2 = (7/5)^y or 2 = (1.4)^y

Solve for y = ln 2 / ln 1.4

Substitute into the first equation for y
4^(x) = 5 ^(ln 2/ ln 1.4)

x ln 4 = (ln2/ln1.4) (ln 5); change ln 4 to ln 2^2 = 2 ln 2

x = (ln 1.4) (ln 5)/(ln 2)