Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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} $

I then had:

$\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!

share|improve this question
2  
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

4 Answers 4

up vote 6 down vote accepted

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

so equation is

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

solving these:

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

share|improve this answer

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

share|improve this answer

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

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

share|improve this answer

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 $x=\dfrac {3}{22}$.

share|improve this answer
    
Please use LaTeX –  gt6989b Jun 7 '13 at 18:03

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.