Logarithmic simultaneous equations

**Hellbent** · November 28th 2010, 03:13 PM

Hi,

I'm unsure of how to approach this.

Solve the simultaneous equations:

$\log x + 2 \log y = 3$
$&x^2y = 125$

**pickslides** · November 28th 2010, 03:18 PM

$y = \frac{125}{x^2}$

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

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

This log base e?

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

You got it from here?

**Plato** · November 28th 2010, 03:26 PM

Assume that $\log$ is base ten.
The first equation becomes $xy^2=10^3$ .

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

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

Can you solve that?

**Hellbent** · November 28th 2010, 04:25 PM

Originally Posted by pickslides

$y = \frac{125}{x^2}$

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

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

This log base e?

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

You got it from here?

Base 10.

I've tried:

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

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

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

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

$\displaystyle \frac{15625}{3} = x$

**Hellbent** · November 28th 2010, 04:29 PM

Originally Posted by Plato

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

I got it now.

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

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

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

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

$xy^2=10^3$

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

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

$5^6x^{-3}=10^3$

**Plato** · November 28th 2010, 04:46 PM

Originally Posted by Hellbent

I don't get it.

Now there is a true and honest statement.

**Hellbent** · November 28th 2010, 05:44 PM

Originally Posted by Plato

Now there is a true and honest statement.

That doesn't help.

**pickslides** · November 28th 2010, 06:04 PM

Well ,my working was base e.

So with base 10

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

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

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

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

$\sqrt[3]{\frac{125^2}{10^3}}= x$

**Hellbent** · November 28th 2010, 06:19 PM

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

$x=2.5$

Substituting x in $x^2y=125$

$(2.5)^2y=125$

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

$y=20$

**Hellbent** · November 28th 2010, 07:26 PM

Originally Posted by Plato

Assume that $\log$ is base ten.
The first equation becomes $xy^2=10^3$ .

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

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

Can you solve that?

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

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

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

$\displaystyle 125=8x^3$

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

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

$\displaystyle 2.5=x$

**Soroban** · November 28th 2010, 07:28 PM

Hello, Hellbent!

Another approach . . .

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

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

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

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

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

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

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