Question

Find all roots of the equation:

Answer 1

`\left| \log_7(x^8)\cfrac{}{} \right|~~ - ~~\left( ~~ \log_(49)(x^2) ~~ \right)^2 = 7 \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \left| \log_7(x^8)\cfrac{}{} \right|\implies \pm\log_7(x^8)\implies \pm 8\log_7(x) \\\\[-0.35em] ~\dotfill\\\\ \left( ~~ \log_(49)(x^2) ~~ \right)^2\implies \left( ~~ \cfrac{\log_7(x^2)}{\log_7(49)} ~~ \right)^2\implies \left( ~~ \cfrac{\log_7(x^2)}{\log_7(7^2)} ~~ \right)^2`

`\left( ~~ \cfrac{\log_7(x^2)}{2} ~~ \right)^2\implies \left( ~~ (1)/(2)\log_7(x^2) ~~ \right)^2\\\\\\ \left( ~~ \log_7(x^{2\cdot (1)/(2)}) ~~ \right)^2 \implies \left( ~~ \log_7(x) ~~ \right)^2 \\\\[-0.35em] ~\dotfill\\\\ \left| \log_7(x^8)\cfrac{}{} \right|~~ - ~~\left( ~~ \log_(49)(x^2) ~~ \right)^2=7 \implies \pm 8\log_7(x)~~ - ~~\left( ~~ \log_7(x) ~~ \right)^2=7 \\\\[-0.35em] ~\dotfill\\\\ \textit{now let's make }\hspace{5em}\log_7(x)=Z \\\\[-0.35em] ~\dotfill`

`\mp 8Z-Z^2=7\implies 0=Z^2\pm 8Z+7\implies 0= \begin{cases} Z^2+ 8Z+7\\ Z^2 - 8Z+7 \end{cases} \\\\[-0.35em] ~\dotfill\\\\ 0=Z^2+8Z+7\implies 0=(Z+7)(Z+1)\implies 0=[\log_7(x)+7][\log_7(x)+1] \\\\\\ 0=Z^2-8Z+7\implies 0=(Z-7)(Z-1)\implies 0=[\log_7(x)-7][\log_7(x)-1]`

now, let's process for each case to get its roots

`0=\log_7(x)+7\implies -7=\log_7(x)\implies 7^(-7)=7^(\log_7(x))\implies \boxed{7^(-7)=x} \\\\\\ 0=\log_7(x)+1\implies -1=\log_7(x)\implies 7^(-1)=7^(\log_7(x))\implies \boxed{7^(-1)=x} \\\\\\ 0=\log_7(x)-7\implies 7=\log_7(x)\implies 7^(7)=7^(\log_7(x))\implies \boxed{7^(7)=x} \\\\\\ 0=\log_7(x)+1\implies 1=\log_7(x)\implies 7^(1)=7^(\log_7(x))\implies \boxed{7=x}`