I'm stuck in a logarithm question: $4^{y+3x} = 64$ and $\log_x(x+12)- 3 \log_x4= -1$

Question

If $4^{y+3x} = 64$ and $\log_x(x+12)- 3 \log_x4= -1$ so $x + 2y= ?$

I've tried this far, and I'm stuck $$\begin{align}4^{y+3x}&= 64 \\ 4^{y+3x} &= 4^3 \\ y+3x &= 3 \end{align}$$

$$\begin{align}\log_x (x+12)- 3 \log_x 4 &= -1 \\ \log_x (x+12)- \log_x 4^3 &= -1 \\ \log_x(x+12)- \log_x 64 &= -1 \end{align}$$

then I substituted $4^{y+3x} = 64$ $\log_x (x+12) - \log_x 4^{y+3x} = -1$

I don't know what should I do next. any ideas?

Answer 1

You're right up to $y+3x=3$.

Now consider the other statement $\log_x(x+12)-3\log_x 4=-1$

$\log_x{x+12 \over 64 }=-1$

${x+12 \over 64 }={1 \over x}$

Answer 2

Now you have

\begin{equation} \log_x\left(\frac{x+12}{64}\right)=-1 \end{equation}

Therefore

\begin{equation} \frac{x+12}{64}=x^{-1} \end{equation}

Which leads to

\begin{equation} x^2-12x-64=0 \end{equation}

Which can be factored

\begin{equation} (x+16)(x-4)=0 \end{equation}

But of course $x$ cannot equal $-16$ so it must equal $4$.

I believe you can take the problem from here.

Answer 3

Consider $$\log_x(x+12)- 3 \log_x(4)= -1$$ and change base to get $$\frac{\log (x+12)}{\log (x)}-\frac{3 \log (4)}{\log (x)}=-1$$ Assuming $x\neq 1$, multiply each term by $\log (x)$ to get $$\log (x+12)-3 \log (4)=-\log (x)$$ thatt is to say $$\log (x+12)+\log(x)=\log(4^3)=\log(64)$$ $$\log\left(x(x+12)\right)=\log(64)$$ $$x(x+12)=64\implies x^2+12x-64=0$$ the roots of which geing $-16$ (to be discarded since $x$ must be positive) and $x=4$ which is then the only root.

Then $y$ from what you already established.

Answer 4

$4^{y+3x} = 64$

$\log_4 ({4^{y+3x}}) = \log_4 (64)$

$(y+3x)\log_4 {4} = 3$

$y + 3x = 3$

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$\log_x (x+12) - 3\log_x (4) = -1$

$\log_x (\frac{x+12}{4^3}) = -1$

$\frac{x+12}{64} = x^{-1}$

$x^2 + 12x - 64 = 0$

$(x+16)(x-4) = 0$

$x = 4$, as $x > 1$ for $\log_x (x+12) - 3\log_x (4) = -1$

(Why must the base of a logarithm be a positive real number not equal to 1?)

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$y + 3(4) = 3$

$y = -9$

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$4 + 2(-9) = -14$