If $x$ and $y$ are positive real numbers such that $\log_{x}\left(x^{2}+12\right)=4$ and $3 \log _{y} x=1$, then $x+y$ equals

We start with: $\log_{x}(x^{2}+12)=4$

When we have $\log_b(a) = c$, this means $b^c = a$. This is the fundamental definition of logarithms.

Applying this to our equation:

Base: $x$

Exponent: $4$

Result: $x^2 + 12$

So: $x^4 = x^2 + 12$

$x^4 - x^2 - 12 = 0$

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Notice this looks like a quadratic equation if we think of $x^2$ as our variable. Let's substitute $u = x^2$:

$u^2 - u - 12 = 0$

We need two numbers that multiply to $-12$ and add to $-1$. These are $-4$ and $+3$.

$(u - 4)(u + 3) = 0$

$(x^2 - 4)(x^2 + 3) = 0$

This gives us: $x^2 = 4$ or $x^2 = -3$

Since $x^2$ cannot be negative for real numbers, we have $x^2 = 4$, so $x = 2$ or $x = -2$.

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In $\log_x$, the base $x$ must be positive and not equal to 1.

Since $x$ appears as the base of a logarithm, we must have $x > 0$.

Therefore: $x = 2$

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Now we use: $3 \log_{y} x = 1$

$\log_{y} x = \tfrac{1}{3}$

$\log_y x = \tfrac{1}{3}$ means $y^{1/3} = x$

Since $x = 2$: $y^{1/3} = 2$

To solve $y^{1/3} = 2$, we cube both sides:

$(y^{1/3})^3 = 2^3$

$y = 8$

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$x + y = 2 + 8 = 10$

When solving logarithmic equations, always remember to convert between logarithmic and exponential forms using the definition: $\log_b(a) = c \iff b^c = a$. Also, check that your solutions satisfy the domain restrictions (positive bases for logarithms).