# logarithm equation!! need help solving...

• Dec 22nd 2010, 09:08 PM
sorkii
logarithm equation!! need help solving...
The questions asks me to find the relationship between x and y by removing the 'log10' in the equation

2log10y + 3log10x = 2

The answer is supposedly y = 10x^3/2

I keep getting up to log10y = 3log10x but don't know how to continue on from there...

• Dec 22nd 2010, 09:11 PM
dwsmith
$\displaystyle log_{10}y^2+log_{10}x^3=2\Rightarrow log_{10}(y^2*x^3)=2\Rightarrow 10^{log_{10}(y^2*x^3)}=10^2\Rightarrow\cdots$
• Dec 22nd 2010, 09:27 PM
sorkii
sorry, i get your answer right up to the last step.....how do i get from '10^log10(y^2*x^3) = 10^2' to y = ....??
• Dec 22nd 2010, 09:29 PM
dwsmith
$\displaystyle 10^{log_{10}x}=x$

It is the inverse.

$\displaystyle e^{ln(x)}=x$ Same concept. This is in base e.
• Dec 22nd 2010, 09:35 PM
sorkii
um...so =>

y^2*x^3 = 10^2
y^2 = 10^2/x^3

so y=√(〖10〗^2/x^3 )

?????

im really sorry...
• Dec 22nd 2010, 09:37 PM
dwsmith
Quote:

Originally Posted by dwsmith
$\displaystyle log_{10}y^2+log_{10}x^3=2\Rightarrow log_{10}(y^2*x^3)=2\Rightarrow 10^{log_{10}(y^2*x^3)}=10^2\Rightarrow\cdots$

$\displaystyle y^2*x^3=10^2\Rightarrow y^2=\frac{10^2}{x^3}\Rightarrow y=\frac{10}{x^{3/2}}$
• Dec 22nd 2010, 09:40 PM
sorkii
okay, i get it now (thanks!!)
just one problem...the textbook says the answer is 10*x^3/2 not 10/x^3/2
is that the books mistake or am i missing something?
• Dec 22nd 2010, 09:41 PM
dwsmith
It should say then $\displaystyle y=10x^{-3/2}$
• Dec 22nd 2010, 09:43 PM
sorkii