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

1. ## 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...

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

3. 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 = ....??

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

It is the inverse.

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

5. um...so =>

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

so y=√(〖10〗^2/x^3 )

?????

im really sorry...

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

7. 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?

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

9. probably....thanks again for your time!