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

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 2

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

Answer 5

Solve the Equation System

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

Step 1: Solving for $x$

The key trick is apply the power-transform $x^h$ to both sides of equation $II$ which will yield a quadratic equation with elementary solutions. Here are the algebrarics:

1. Multiply the equation with $-1$: $3\log_x 4-\log_x(x+12)=1$
2. Take both sides of the equation to the power of $x$, i.e. $x^{LHS}=x^{RHS}$ and solve for $x$ using the properties of the logarithm function:

\begin{align} &&x^{3\log_x(4)-\log_x(x+12)}&=x^1\\ \leftrightarrow&&\frac{x^{3\log_x(4)}}{x^{\log_x(x+12)}}&=x \end{align} Observe that applying a logarithmic term with (its) inverse (i.e. power function) we obtain the identity function, that is $h^{\log_h (s)}=s$. Thus, the left handside of the equation also satisfies $\frac{x^{3\log_x(4)}}{x^{\log_x(x+12)}}=\frac{4^3}{x+12}$. Thus, we finally have

$$ \frac{4^3}{x+12x}=x $$

3. Pushing all terms to the right-hand side, we get the quadratic equation $$ x^2+12x-64=0. $$ The roots of a quadratic equation can be found using the $p/q$ formula, i.e.

$$ x^*=-6\pm\sqrt{100}=-6\pm 10 $$

Notice that we have two solutions, for this equation $x^*\in\{4,-16\}$. But negative values of $x$ can be excluded as the expressions above, specifically $\log_x$ are not defined for $x<0$. Hence, the solution must be

$$ x^*=4 $$

Step 2: Solving for $y$

Substitute $x^*$ into equation $I$:

\begin{align} &&4^{y+3*x^*}&=64\\ &&4^{y+12}&=64 \end{align} Take the $log_4$ to on both sides of the equation,

\begin{align} y+12=3 \end{align}

This gives $$ y^*=-9. $$