The solution to the equation {log _7}{log _5} | vedclass

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Question

(C) Given the equation: ${\log _7}{\log _5}(\sqrt {{x^2} + 5 + x} ) = 0$ Applying the definition of logarithm,${\log _b}(a) = c \implies a = b^c$: ${\

Vedclass · Accepted Answer

$x = 4$

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(C) Given the equation: ${\log _7}{\log _5}(\sqrt {{x^2} + 5 + x} ) = 0$ Applying the definition of logarithm,${\log _b}(a) = c \implies a = b^c$: ${\log _5}(\sqrt {{x^2} + 5 + x} ) = 7^0 = 1$ Applying the definition again: $\sqrt {{x^2} + 5 + x} = 5^1 = 5$ Squaring both sides: ${x^2} + x + 5 = 25$ ${x^2} + x - 20 = 0$ Factoring the quadratic equation: $(x + 5)(x - 4) = 0$ So,$x = -5$ or $x = 4$. Checking the domain: The argument of the outer logarithm must be positive,and the argument of the inner logarithm must be positive. For $x = 4$: $\sqrt{16 + 5 + 4} = \sqrt{25} = 5 > 0$. ${\log _5}(5) = 1 > 0$. This is valid. For $x = -5$: $\sqrt{25 + 5 - 5} = \sqrt{25} = 5 > 0$. ${\log _5}(5) = 1 > 0$. This is also valid. However,looking at the options provided,$x = 4$ is the correct choice.

The solution to the equation ${\log _7}{\log _5}(\sqrt {{x^2} + 5 + x} ) = 0$ is:

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