logarithm equation!! need help solving...

**sorkii** · December 22nd 2010, 09:08 PM

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...

Please help!

**dwsmith** · December 22nd 2010, 09:11 PM

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

**sorkii** · December 22nd 2010, 09:27 PM

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 = ....??

**dwsmith** · December 22nd 2010, 09:29 PM

$\displaystyle 10^{log_{10}x}=x$

It is the inverse.

$\displaystyle e^{ln(x)}=x$ Same concept. This is in base e.

**sorkii** · December 22nd 2010, 09:35 PM

um...so =>

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

so y=√(〖10〗^2/x^3 )

?????

im really sorry...

**dwsmith** · December 22nd 2010, 09:37 PM

Originally Posted by dwsmith

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

$\displaystyle y^2*x^3=10^2\Rightarrow y^2=\frac{10^2}{x^3}\Rightarrow y=\frac{10}{x^{3/2}}$

**sorkii** · December 22nd 2010, 09:40 PM

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?

**dwsmith** · December 22nd 2010, 09:41 PM

It should say then $\displaystyle y=10x^{-3/2}$

**sorkii** · December 22nd 2010, 09:43 PM

probably....thanks again for your time!

Thread: logarithm equation!! need help solving...

LinkBack

Thread Tools

Display

logarithm equation!! need help solving...

Similar Math Help Forum Discussions

solving logarithm

Help with solving this logarithm

Solving for x in a logarithm

Need help solving this logarithm

Solving a logarithm equation

Search Tags