Question 46

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

Solution

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.

Video Solution

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