Question

Please Solve the question given in the attachment.........

Please don't answer if you don't know.......

Answer 1

Given 2^x = 3^y = 6^z.

Apply log on both sides, we get

log(2^x) = log(3^y) = log(6^z).

We know that log (a^b) = b log a

x log 2 = y log 3 = z log 6 = k ---- Some constant value


Now, 

x log 2 = k

x = k/log 2  ------ (1)


y log 3 = k

y = k/log 3   ------- (2)


z log 6 = k

z = k/log 6    ----------- (3)


Given Equation is:

$\frac{1}{x} + \frac{1}{y} - \frac{1}{z} = \frac{1}{ \frac{k}{log 2}} + \frac{1}{ \frac{k}{log 3} } - \frac{1}{ \frac{k}{log 6} }$   


                        = $\frac{log 2 + log 3 - log 6}{k}$


We know that log a + log b = log ab

                                                  = $\frac{log(2 * 3) - log(6)}{k}$

                                                  = $\frac{log 6 - log 6}{k}$

                                                  = 0.



Hope this helps!

Answer 2

Hi,

Please see the attached file!

Thanks