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Question Number 198489 by sonukgindia last updated on 21/Oct/23

Answered by cortano12 last updated on 21/Oct/23

$${\mathrm{x}}^{16{\left(\log {}_5\mathrm{x}\right)}^3-68\log {}_5\mathrm{x}}=5^{-16}$$$$\mathrm{let}\log {}_5\mathrm{x}=\mathrm{t}$$$${\mathrm{x}}^{16{\mathrm{t}}^3-68\mathrm{t}}=5^{-16}$$$$({16\mathrm{t}}^3-68\mathrm{t})\mathrm{t}=-16$$$${4\mathrm{t}}^4-{17\mathrm{t}}^2+4=0$$$$({4\mathrm{t}}^2-1)({\mathrm{t}}^2-4)=0$$$$\{\begin{array}{l}\mathrm{t}=\log {}_5\mathrm{x}=\pm \frac{1}{2}\Rightarrow \mathrm{x}=\sqrt{5},\mathrm{x}=\frac{1}{\sqrt{5}}\\ \mathrm{t}=\log {}_5\mathrm{x}=\pm 2\Rightarrow \mathrm{x}=5^2;\mathrm{x}=\frac{1}{25}\end{array}$$$$$$

Commented by Rasheed.Sindhi last updated on 21/Oct/23

$${\mathrm{x}}^{16{\mathrm{t}}^3-68\mathrm{t}}=5^{-16}\stackrel{???}{\Rightarrow }({16\mathrm{t}}^3-68\mathrm{t})\mathrm{t}=-16$$

Commented by cortano12 last updated on 21/Oct/23

$$\log {}_5\left({\mathrm{x}}^{{16\mathrm{t}}^3-68\mathrm{t}}\right)=\log {}_5\left(5^{-16}\right)$$$$({16\mathrm{t}}^3-68\mathrm{t})\mathrm{t}=-16$$

Commented by Frix last updated on 21/Oct/23

$$(4t^2-1)(t^2-4)=0$$

Answered by Frix last updated on 21/Oct/23

$$x={\mathrm{e}}^t$$$${\mathrm{e}}^{\frac{16t^4}{{\ln }^{3}5}-\frac{68t^2}{\ln 5}}=5^{-16}$$$$\frac{16t^4}{{\ln }^{3}5}-\frac{68t^2}{\ln 5}=-16\ln 5$$$$t^4-\frac{{17\ln }^{2}5}{4}{t}^2+{\ln }^{4}5=0$$$$$$$$t=\pm \frac{\ln 5}{2}\vee t=\pm 2\ln 5$$$$x=\frac{1}{25}\vee x=\frac{\sqrt{5}}{5}\vee x=\sqrt{5}\vee x=25$$

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