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$2^{x}=5^{y}=100^{z}$

Find $z$ in terms of $y$ and $x$.


The term $z$ should be a function of $x$ and $y$, i.e.: $z(x,y)$.

All I could get were recursive attempts.

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  • $\begingroup$ Hint: if $100^z = a$, what is $z$ in terms of $a$ ? $\endgroup$ – Fabrice NEYRET Nov 2 '15 at 13:44
  • $\begingroup$ Take the logarithm of both sides. $\endgroup$ – Aretino Nov 2 '15 at 13:44
  • $\begingroup$ $\log _{100}\left( a \right)=z$ ? $\endgroup$ – inspd Nov 2 '15 at 13:45
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Suppose $z$ is non-zero.

$100^z=a\rightarrow\log_a100=\frac{1}{z}$

Likewise, we have $\log_a2=\frac{1}{x}$ and $\log_a5=\frac{1}{y}$.

Hence, $\frac{1}{z}=\log_a100=2\log_a2+2\log_a5=\frac{2}{x}+\frac{2}{y}$.

Rewriting gives: $z=\frac{xy}{2(x+y)}$.

Taking into account the case when $z=0$, it follows that:

$$z=\begin{cases}\frac{xy}{2(x+y)}&\text{if $x\neq0$},\\0&\text{otherwise}\end{cases}$$

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  • $\begingroup$ this is a nice approach, thanks $\endgroup$ – inspd Nov 2 '15 at 14:32
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HINT: $100^z=10^{2z}=2^{2z} \cdot 5^{2z}$

Hence the relation becomes $2^x=5^y=2^{2z} \cdot 5^{2z}$

EDIT: $$2^x=5^y=2^{2z} \cdot 5^{2z}=a$$

Therefore, we have $$x\log 2=y\log 5=2z(\log 2 + \log 5)$$ $$\frac{x-2z}{2z}=\frac{\log 5}{\log 2}$$ and $$\frac{2z}{y-2z}=\frac{\log 5}{\log 2}$$ So we have $$\frac{x-2z}{2z}=\frac{2z}{y-2z}$$ or,$$xy-2z(x+y)+4z^2=4z^2$$ or, $$z=\frac{xy}{2(x+y)}$$

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  • $\begingroup$ can you confirm this: $$z=\frac{xy}{2\left( x+y \right)}$$ $\endgroup$ – inspd Nov 2 '15 at 14:01
  • $\begingroup$ I can confirm. Currently writing a complete answer. $\endgroup$ – Element118 Nov 2 '15 at 14:16
  • $\begingroup$ Updated my answer. Sorry for being late. See if it is okay. $\endgroup$ – SchrodingersCat Nov 2 '15 at 14:40
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Taking the logarithm, to any base, of each of those, x log(2)= y log(5)= z log(100). z= xlog(2)/log(100)= y log(5)/log(100).

If you se the common logarithm, base 10, log(100)= 2 so those become z= xlog(2)/2= y log(5)/2.

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  • $\begingroup$ it should be in terms of $x$ and $y$, i.e.: $z(x,y)$. I'll specify more in the question $\endgroup$ – inspd Nov 2 '15 at 13:53

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