# Solve these equations simultaneously

Solve these equations simultaneously: \eqalign{ & {8^y} = {4^{2x + 3}} \cr & {\log _2}y = {\log _2}x + 4 \cr}

I simplified them first:

\eqalign{ & {2^{3y}} = {2^{2\left( {2x + 3} \right)}} \cr & {\log _2}y = {\log _2}x + {\log _2}{2^4} \cr}

\eqalign{ & 3y = 4x + 6 \cr & y = x + 16 \cr}

Solving:

\eqalign{ & 3\left( {x + 16} \right) = 4x + 6 \cr & 3x + 48 = 4x + 6 \cr & x = 42 \cr & y = \left( {42} \right) + 16 \cr & y = 58 \cr}

This is the wrong answer, I would like to understand where I went wrong so I dont make the same mistake again, your help is greatly appreciated, thanks!

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Note that $\log_2 x + \log_2 (2^4) = \log_2 (16x)$ so your second equation should be $y = 16x$. – Suugaku Jun 7 '13 at 16:53
@GitGud I think the OP is trying to combine what's inside the logs. They used the fact that $4 = \log_2 (2^4)$ to get everything in terms of $\log_2$. – Suugaku Jun 7 '13 at 16:56
Thanks! I've got it now! – seeker Jun 7 '13 at 16:57

$$\log _2y = {\log _2}x + \log_22^4$$ $$\log m+\log n=\log (mn)$$ $$\log _2y = \log _2({x\cdot 2^4})$$ $$y=16x$$ this will be second eqn

so equation is

$3y = 4x + 6\;\;$ and $y=16x$

solving these:

$3\times16x = 4x + 6\implies44x=6\implies x=\dfrac3{22 }\;\;,y=\dfrac{24}{11}$

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$$\log_2 y = (\log_2 x) + 4$$ $$2^{\log_2 y} = 2^{\log_2 x}\cdot 2^4$$ $$y = x\cdot 16$$ You added where you needed to multiply.

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Recall: $\quad\log_2(a) + \log_2(b) = \log_2(ab)$. So

$$\log_2 x + \log_2 (2^4) = \log_2(x) + \log_2(16) = \log_2 (16x)$$

So your system of equations should be

$$3y = 4x + 6$$ $$y = 16 x$$

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The equation wil be $\log_2 y= \log_2 x + \log_2 2^4$. so $y$ will then be equal to $16x$. So $4x$ will be $\dfrac y4$. Substituting,so $3y=\dfrac y4+6$

$\dfrac {11y}4= 6$

So $y=\dfrac {24}{11}$. And $x=\dfrac {3}{22}$.

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Please use LaTeX – gt6989b Jun 7 '13 at 18:03