# Thread: exponentials and logarithms

1. ## exponentials and logarithms

Show using:

3^x = 4^y = 12^z,

that z = xy/(x+y)

Tried looking at 3^x = 4^y and taking log base 12 to eliminate one of the variables, ending up with x=ylog base 3 (4).

then I looked at 4^y = 12^z, took log base 12 as well, and ended up with z= y*log base 12 (4).

Plugged x and y into xy/(x+y), but couldnt get it to equal z

had a fiddle about with it, but I can't seem to prove it. maybe, im thinking too hard, anybody else got ideas?

2. $\displaystyle 3^x = 4^y = 12^z = k$

$\displaystyle 3 = k^1/x, 4 = k^1/y and 12 = k^1/z$

12 = 3*4

Substitute the values and simplify.

3. $\displaystyle 3^x = 4^y = 12^z = k$

$\displaystyle 3 = k^{1/x}, 4 = k^{1/y} and 12 = k^{1/z}$

12 = 3*4

Substitute the values and simplify.

4. from

$\displaystyle 12^z = 4^y$ find z value

$\displaystyle z = \log_{12} 4^y$

$\displaystyle z = y \log_{12} 4$ same thing from $\displaystyle 12^z = 3^x$

you will have

$\displaystyle z = y \log_{12} 4$
$\displaystyle z = x \log_{12} 3$

multiply first one with x and the second with y

$\displaystyle zx = yx \log_{12} 4$
$\displaystyle zy = yx \log_{12} 3$

$\displaystyle zx + zy = yx( \log_{12} 4 + \log_{12} 3 )$

$\displaystyle z (x+y) = yx (\log_{12} 3(4) )$

$\displaystyle z = \frac{xy}{x+y}$

5. or u can find y,x values with respect to z then sub them in xy/(x+y) u should get z

6. thanks guys.

got it!