If $2^x=3^y=6^{-z}$ then prove that:$ \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0$

Question

If $$2^x=3^y=6^{-z}$$ and $x,y,z \neq 0 $ then prove that:$$ \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0$$

I have tried starting with taking logartithms, but that gives just some more equations.

Any specific way to solve these type of problems?

Any help will be appreciated.

Answer 1

$2^x=3^y=6^{-z}=k $ say, then $2= k^{1\over x},3=k^{1\over y},6=k^{-1\over z}$ now can you go on?

then $k^{-1\over z}=6=2\times 3 = k^{1\over x}\times k^{1\over y}=k^{{1\over x}+{1\over y}}$

Answer 2

As $\log_aa=1$ and $\log_a(bc)=\log_ab+\log_ac,$

applying logarithm wrt $2,$

$x=y\log_23=-z\log_26=-z(1+\log_23)$

$x=y\log_23\implies \log_23=\frac xy$

and put this value of $\log_23$ in $\displaystyle x=-z(1+\log_23)$

to eliminate $\log_23$ and simplify.

Answer 3

Take logarithms (any base) to obtain $$x\log 2= y\log 3=-z\log 6=-z\log 3-z\log 2$$

Then note that $(y+z)\log 3=-z\log 2$ and $x\log 2 = y\log 3$

Multiply left- and right- hand sides to obtain $$(xy+yz)\log 3 \log 2=-yz\log 2\log 3$$

Whence $xy+yz+zx=0$

Note that we have done no division so far, except by the non-zero term $\log 2\log 3$, and also that $x=y=z=0$ is a solution. If $xyz\neq 0$ we can divide by $xyz$ to obtain the required equation.

Answer 4

$$2^x = 3^y = 6^{-z} = k $$

so$$x = \log_2k$$ $$ y = \log_3k$$ $$z= -\log_6k$$

so $$\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \log_k2 + log_k3 -\log_k6$$ $$=\log_k{\frac{2\times3}{6}}$$ $$=0$$