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If $$x$$ and $$y$$ are positive real numbers such that $$\log_{x}(x^2 + 12) = 4$$ and $$3 \log_{y} x = 1$$, then $$x + y $$ equals
Given, $$\log_{x}(x^2 + 12) = 4$$
=> $$x^2+12=x^4$$
=> $$x^4-x^2-12=0$$
=> $$x^4-4x^2+3x^2-12=0$$
=> $$x^2\left(x^2-4\right)+3\left(x^2-4\right)=0$$
=> $$\left(x^2-4\right)\left(x^2+3\right)=0$$ => since, x is a positive real number (given) => x = 2.
Now, Given $$3 \log_{y} x = 1$$
=> $$\log_yx=\frac{1}{3}$$
=> $$x=y^{\frac{1}{3}}$$
=> $$y=x^3$$ => y = 8.
=> x + y = 2 + 8 = 10.
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