Find the solution of the equation log_{7}(log | vedclass

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Question

(C) Given the equation $\log_{7}(\log_{5}(\sqrt{x^2 + x + 5})) = 0$. Since $\log_{7}(1) = 0$,we have $\log_{5}(\sqrt{x^2 + x + 5}) = 1$. This implies

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$x = 4$

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(C) Given the equation $\log_{7}(\log_{5}(\sqrt{x^2 + x + 5})) = 0$. Since $\log_{7}(1) = 0$,we have $\log_{5}(\sqrt{x^2 + x + 5}) = 1$. This implies $\sqrt{x^2 + x + 5} = 5^1 = 5$. Squaring both sides,we get $x^2 + x + 5 = 25$. Rearranging the terms,$x^2 + x - 20 = 0$. Factoring the quadratic equation,$(x + 5)(x - 4) = 0$. Thus,$x = 4$ or $x = -5$. Checking the domain,for $\log_{5}(\sqrt{x^2 + x + 5})$ to be defined,$\sqrt{x^2 + x + 5} > 0$,which is satisfied for both values. However,the original equation is $\log_{7}(\log_{5}(\sqrt{x^2 + x + 5})) = 0$. For $x = 4$,$\sqrt{16+4+5} = \sqrt{25} = 5$,and $\log_{5}(5) = 1$,$\log_{7}(1) = 0$. For $x = -5$,$\sqrt{25-5+5} = \sqrt{25} = 5$,and $\log_{5}(5) = 1$,$\log_{7}(1) = 0$. Given the options,$x = 4$ is the correct choice.

Find the solution of the equation $\log_{7}(\log_{5}(\sqrt{x^2 + x + 5})) = 0$.

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