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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.

  • 1
    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](http://en.wikipedia.org/wiki/Exponentiation#Identities_and_properties).2011-09-20

3 Answers 3

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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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    @Gareth By the way, sometimes when we manipulate equations (e.g., eliminate logs by e$x$ponentiating), 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.2011-09-20
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$2^y=2^{2x} \Rightarrow y=2x$, 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 :)2012-12-15
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From equation $y^2 =2x$ $y=\sqrt{2x}$

Substitute values of $y$ in the equation above.

If you work it smartly, you will find that $y=1, x=0.5$

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    Hi, the answers are much more readable if you use `\sqrt{2x}` (this results in $\sqrt{2x}$) instead of 'square root of 2x'. More the other tricks, follow [this link](https://math.meta.stackexchange.com/q/5020/73561).2017-04-29