3
$\begingroup$

Question

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.

Thanks!

$\endgroup$

2 Answers 2

Reset to default
1
$\begingroup$

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

$$\log_2x+4=\log_2x+\log_216=\log_216x$$

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

$\endgroup$
1
$\begingroup$

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

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

$\endgroup$
5
  • $\begingroup$ 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, 2015 at 1:44
  • 1
    $\begingroup$ @baxx Raising both sides to the power of $2$. $\endgroup$
    – MathMajor
    May 13, 2015 at 1:45
  • 2
    $\begingroup$ @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, 2015 at 1:47
  • $\begingroup$ There's a mistake in the last step. $8^{16x}\ne4^{32x}$ $\endgroup$
    – Mike
    May 13, 2015 at 1:49
  • 1
    $\begingroup$ @Mike fixed $$$$ $\endgroup$
    – Teoc
    May 13, 2015 at 1:55

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .