Solve giving your answers as exact fractions, the simultaneous equations :

$$8^y = 4^{2x + 3} \tag{1}$$

$$\log_2 y = \log_2x + 4 \tag{2}$$

I think that the RHS of eq 1 can be split up, I'm hoping that something will fall into place after trying at least.

$$4^{2x + 3} = 4^3(4^{2x})$$

I'm not sure if having $4^{2x}$ on it's own helps here... The problem is that I'm not too sure what I'm trying to achieve with this problem. I get that Its a simultaneous with logarithms. I want to find a term that I can substitute into one of the equations, or some thing that I can subtract.

I can't do $8^y - \log_2 y $ though.

I want to try and get the top equations into $\log_2$ form so that maybe I can start substituting things about, but I can't seem to do that either.

I'm not actually sure how I would express $8^y$ in $\log_2$. It's base 8 isn't it.

Would be cool to get any advice on this, don't really know what to do with it.



For the second equation, the right side can be rewritten as


at which point you can eliminate the logarithms. As for the first equation, I'd recommend expressing both sides as a power of $2$.


$$\log_2 y=\log_2x+4$$ $$y=2^4x$$ $$y=16x$$ Plug into first equation

$$8^{16x}=4^{2x+3}$$ $$4^{24x}=4^{2x+3}$$ $$\vdots$$ Can you take it from there?

  • $\begingroup$ thanks @MathNoob - I'm actually unsure of the process that got rid of $\log_2$ on the first part. If you could explain that / provide me a link to somewhere that does that'd be ace :) $\endgroup$ – baxx May 13 '15 at 1:44
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    $\begingroup$ @baxx Raising both sides to the power of $2$. $\endgroup$ – MathMajor May 13 '15 at 1:45
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    $\begingroup$ @baxx $$\log_2 y=\log_2 x +4$$ $$2^{\log_2 y}=2^{\log_2 x +4}$$ $$y=2^{\log_2 x}2^4$$ $$y=16x$$ $\endgroup$ – Teoc May 13 '15 at 1:47
  • $\begingroup$ There's a mistake in the last step. $8^{16x}\ne4^{32x}$ $\endgroup$ – Mike May 13 '15 at 1:49
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    $\begingroup$ @Mike fixed $$$$ $\endgroup$ – Teoc May 13 '15 at 1:55

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