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### Solve for $x\in R$, $2^x+4^x=8^x$

Using the substitution $2^x=t$, we have

$t+t^2=t^3$

$\Rightarrow {t^3} - {t^2} - t = 0$

Or, $t({t^2} - t - 1) = 0$

$t \ne 0$ as $t=2^x>0$

So, ${t^2} - t - 1 = 0$

Or, $t = \frac{{1 \pm \sqrt {1 + 4} }}{2} = \frac{{1 \pm \sqrt 5 }}{2}$

Since t > 0, $t = \frac{{1 + \sqrt 5 }}{2}=2^x$

So, $x\log 2 = \log \left( {\frac{{1 + \sqrt 5 }}{2}} \right)$

$\Rightarrow x = \frac{{\log \left( {\frac{{1 + \sqrt 5 }}{2}} \right)}}{{\log 2}} = \frac{{\log (1 + \sqrt 5 )}}{{\log 2}} - 1$

### $f(x)=x^6+2x^4+x^3+2x+3 $$\mathop {\lim }\limits_{x \to 1} \frac{{{x^n}f(1) - f(x)}}{{x - 1}} = 44$$n=?$

Let $f(x)=x^6+2x^4+x^3+2x+3,x \in R$. Then the natural number n for which $\mathop {\lim }\limits_{x \to 1} \frac{{{x^n}f(1) - f(x)}}{{x - 1}} = 44$ is _ _ _ _ . Solution Since the limit has $\left[ {\frac{0}{0}} \right]$ form, L.H. Rule is applicable. Thus, $\mathop {\lim }\limits_{x \to 1} n{x^{n - 1}}f(1) - f'(x) = 44$ $\therefore nf(1) - f'(1) = 44$ $\therefore n.9 - ({6.1^5} + {8.1^3} + {3.1^2} + 2.1) = 44$ $\Rightarrow 9n - 19 = 44$ $\Rightarrow n=7$