Question

${3}^{2x + 1} = {3}^{x + 2} + \sqrt{1 - {6.3}^{ x} + {3}^{2(x + 1)} }$
please answer my question
​

Answer 1

Solution :

$\mathrm{Given,\:3^{2x+1}=3^{x+2}+\sqrt{1-6.3^{x}+3^{2(x+1)}}}$

$\to \mathrm{3^{2x+1}=3^{x+2}+\sqrt{1-2.1.3^{x+1}+(3^{x+1})^{2}}}$

$\to \mathrm{3^{2x+1}=3^{x+2}+\sqrt{(1-3^{x+1})^{2}}}$

$\to \mathrm{3^{2x+1}=3^{x+2}+1-3^{x+1}}$

$\to \mathrm{3.3^{2x}=3^{2}.3^{x}+1-3.3^{x}}$

$\to \mathrm{3.(3^{x})^{2}=(9-3).3^{x}+1}$

$\to \mathrm{3z^{2}-6z=1,\:where\:3^{x}=z\:(say)}$

$\to \mathrm{z^{2}-2z+1=1+\frac{1}{3}}$

$\to \mathrm{(z-1)^{2}=(\frac{2}{\sqrt{3}})^{2}}$

$\to \mathrm{(z-1)^{2}-(\frac{2}{\sqrt{3}})^{2}=0}$

$\to \mathrm{(z-1+\frac{2}{\sqrt{3}})(z-1-\frac{2}{\sqrt{3}})=0}$

$\mathrm{Either\:z-1+\frac{2}{\sqrt{3}}=0\:or,\:z-1-\frac{2}{\sqrt{3}}=0}$

$\to \mathrm{z=1-\frac{2}{\sqrt{3}},\:1+\frac{2}{\sqrt{3}}}$

Since z cannot be taken as negative for which the value of $\mathrm{z=3^{x}}$ will be indeterminate.

∴ $\mathrm{z=1+\frac{2}{\sqrt{3}}}$

$\to \mathrm{3^{x}=1+\frac{2}{\sqrt{3}}}$

$\to \mathrm{x\:log_{e}3=log_{e}(1+\frac{2}{\sqrt{3}})}$

$\to \mathrm{x=\frac{log_{e}(1+\frac{2}{\sqrt{3}})}{log_{e}3}}$

$\to \mathrm{x=log_{3}(1+\frac{2}{\sqrt{3}})}$

Hence, the required solution is

x = $\mathrm{log_{3}(1+\frac{2}{\sqrt{3}})}$

or, x = 0.699 ( approximately )