# Logarithmic simultaneous equations

• Nov 28th 2010, 03:13 PM
Hellbent
Logarithmic simultaneous equations
Hi,

I'm unsure of how to approach this.

Solve the simultaneous equations:

$\displaystyle \log x + 2 \log y = 3$
$\displaystyle &x^2y = 125$
• Nov 28th 2010, 03:18 PM
pickslides
$\displaystyle y = \frac{125}{x^2}$

$\displaystyle \log x + \log \left( \frac{125}{x^2}\right)^2 = 3$

$\displaystyle \log \left(x\left( \frac{125}{x^2}\right)^2\right) = 3$

This log base e?

$\displaystyle x\left( \frac{125}{x^2}\right)^2 = e^3$

You got it from here?
• Nov 28th 2010, 03:26 PM
Plato
Assume that $\displaystyle \log$ is base ten.
The first equation becomes $\displaystyle xy^2=10^3$.

The we have from the second $\displaystyle x^2y=5^3$ so $\displaystyle ~y^2=5^6x^{-4}$.

Thus we get $\displaystyle ~5^6x^{-3}=10^3$

Can you solve that?
• Nov 28th 2010, 04:25 PM
Hellbent
Quote:

Originally Posted by pickslides
$\displaystyle y = \frac{125}{x^2}$

$\displaystyle \log x + \log \left( \frac{125}{x^2}\right)^2 = 3$

$\displaystyle \log \left(x\left( \frac{125}{x^2}\right)^2\right) = 3$

This log base e?

$\displaystyle x\left( \frac{125}{x^2}\right)^2 = e^3$

You got it from here?

Base 10.

I've tried:

$\displaystyle \displaystyle \left(\frac{125}{x}\right)^2=e^3$

$\displaystyle \displaystyle \ln\left(\frac{125}{x}\right)^2=\ln e^3$

$\displaystyle \displaystyle 2\ln\frac{125}{x}=3$

$\displaystyle \displaystyle \frac{2\ln125}{3}=x$

$\displaystyle \displaystyle \frac{15625}{3} = x$
• Nov 28th 2010, 04:29 PM
Hellbent
Quote:

Originally Posted by Plato
The we have from the second $\displaystyle x^2y=5^3$ so $\displaystyle ~y^2=5^6x^{-4}$.

I got it now.

$\displaystyle x^2y=5^3$ so $\displaystyle y^2=5^6x^{-4}$

$\displaystyle x^4y^2=5^6$

$\displaystyle \displaystyle y^2=\frac{5^6}{x^4}$

$\displaystyle y^2=5^6x^{-4}$

$\displaystyle xy^2=10^3$

Replacing $\displaystyle ~y^2$ with $\displaystyle ~5^6x^{-4}$

$\displaystyle 5^6x^{-4}x=10^3$

$\displaystyle 5^6x^{-3}=10^3$
• Nov 28th 2010, 04:46 PM
Plato
Quote:

Originally Posted by Hellbent
I don't get it.

Now there is a true and honest statement.
• Nov 28th 2010, 05:44 PM
Hellbent
Quote:

Originally Posted by Plato
Now there is a true and honest statement.

That doesn't help.
• Nov 28th 2010, 06:04 PM
pickslides
Well ,my working was base e.

So with base 10

$\displaystyle x\left( \frac{125}{x^2}\right)^2 = 10^3$

$\displaystyle \frac{125^2x}{x^4}= 10^3$

$\displaystyle \frac{125^2}{x^3}= 10^3$

$\displaystyle \frac{125^2}{10^3}= x^3$

$\displaystyle \sqrt[3]{\frac{125^2}{10^3}}= x$
• Nov 28th 2010, 06:19 PM
Hellbent
$\displaystyle \displaystyle \sqrt[3]{\frac{125^2}{10^3}}= x$

$\displaystyle x=2.5$

Substituting x in $\displaystyle x^2y=125$

$\displaystyle (2.5)^2y=125$

$\displaystyle y=\frac{125}{6.25}$

$\displaystyle y=20$
• Nov 28th 2010, 07:26 PM
Hellbent
Quote:

Originally Posted by Plato
Assume that $\displaystyle \log$ is base ten.
The first equation becomes $\displaystyle xy^2=10^3$.

The we have from the second $\displaystyle x^2y=5^3$ so $\displaystyle ~y^2=5^6x^{-4}$.

Thus we get $\displaystyle ~5^6x^{-3}=10^3$

Can you solve that?

$\displaystyle \displaystyle x^{-3}=\frac{10^3}{5^6}$

$\displaystyle \displaystyle \frac{1}{x^3}=\frac{10^3}{5^6}$

$\displaystyle \displaystyle \frac{1}{x^3}=\frac{8}{125}$

$\displaystyle \displaystyle 125=8x^3$

$\displaystyle \displaystyle \frac{125}{8}=x^3$

$\displaystyle \displaystyle \sqrt[3]{\frac{125}{8}}=x$

$\displaystyle \displaystyle 2.5=x$
• Nov 28th 2010, 07:28 PM
Soroban
Hello, Hellbent!

Another approach . . .

Quote:

$\displaystyle \text{Solve: }\;\begin{array}{cccc}\log x + 2 \log y &=& 3 & [1] \\ x^2y &=& 125 & [2] \end{array}$

From [1]: .$\displaystyle \log x + \log y^2 \:=\:3 \quad\Rightarrow\quad \log(xy^2) \:=\:3$

$\displaystyle \begin{array}{ccccccc} \text{[1] becomes:} & xy^2 &=& 1000 & [3] \\ \text{Divide by [2]:} & x^2y &=& 125 & [2] \end{array}$

We have: .$\displaystyle \displaystyle \frac{xy^2}{x^2y} \:=\:\frac{1000}{125} \quad\Rightarrow\quad \frac{y}{x} \:=\:8 \quad\Rightarrow\quad y \:=\:8x$

Substitute into [2]: .$\displaystyle x^2(8x) \:=\:125 \quad\Rightarrow\quad x^3 \:=\:\frac{125}{8} \quad\Rightarrow\quad \boxed{x \:=\:\tfrac{5}{2}}$

Sustitute into [2]: .$\displaystyle \left(\frac{5}{2}\right)^2y \:=\:125 \quad\Rightarrow\quad \frac{25}{4}y \:=\:125 \quad\Rightarrow\quad \boxed{y \:=\:20}$