Solve these equations simultaneously

Question

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} $

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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!

Answer 1

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

so equation is

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

solving these:

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

Answer 2

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

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

Answer 3

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

Answer 4

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