# Solve simultaneous equations $2\log(y) = \log(2) + \log(x)$ and $2^y = 4^x$

I'm having trouble finding the solution to the following problem.

(1) $2\log(y) = \log(2) + \log(x)$

(2) $2^y = 4^x$

So far all I've managed is:

from (1) simplify using properties of logarithms:

$\log(y^2) = \log(2x)$

$y^2 = 2x$

or from (2)

$2^y = 4^x$

$10^y = \log(y) = 20^x$

neither of which seem to lead to a suitable substitution into the other equation.

tagged as homework as it is obviously homework level, only trouble I don't have a teacher or tutor to ask.

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Your simplification of (2.) is quite wrong. Can you show us how you went from $2^y = 4^{x}$ to $10^y = \log y = 20^x$? – Srivatsan Sep 20 '11 at 19:46
I figured if I multiplied both sides by 5. $2^y.5 = 4^x.5$ I would have $10^y = 20^x$, then $\log y = 20^x$ as you can see that didn't actually help me – Gareth Davis Sep 20 '11 at 19:54
It didn't help you is one thing. In fact, it is wrong. $(2 \cdot 5)^y$ is not $2^y \cdot 5$. It is $2^y \cdot 5^y$. (Also I do not see how you went from $10^y = 20^x$ to $\log y = 20^x$.) You can review the properties of exponentials here. – Srivatsan Sep 20 '11 at 19:57

If $2^y = 4^x$, then since $4=2^2$ we have $$2^y = 4^x = (2^2)^x = 2^{2x},$$ so $y=2x$.

(Alternatively, taking logarithms you have $y\log(2) = x\log(4) = 2x\log(2)$, so $y=2x$).

So now you know that $y^2=2x$ and that $y=2x$. Therefore...

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that's just what i needed, thank you. – Gareth Davis Sep 20 '11 at 19:48
@Gareth By the way, sometimes when we manipulate equations (e.g., eliminate logs by exponentiating), we might end up with what are called "extraneous solutions" (or what you could call "fake solutions"). Remember to always check that each solution is really a solution. – Srivatsan Sep 20 '11 at 19:52

$2^y=2^{2x} \Rightarrow y=2x$, if we plug this into first equation we have $y^2=y \Rightarrow y^2-y=0 \Rightarrow y(y-1)=0$ ,so $y_1=0$ and $y_2=1$ which means that $x_1=0$ and $x_2=\frac{1}{2}$ Since $ln$ isn't defined for $0$ only solution is $(x_2,y_2)=(\frac{1}{2},1)$

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 ok so i have a question - where did the y1 and y2 come from at the end? why is there 2 anwsers. I thought you solved simultaniously by taking away y=2x fromy^2=2x ...that gives y=0 and then i get lost ... pls help :) – karolina Dec 15 '12 at 21:45