Find the value of x x^2 = 16^x please help with workings.


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Question Number 6726 by Tawakalitu. last updated on 16/Jul/16

$F i n d t h e v a l u e o f x$ $x^{2} = 16^{x}$ $p l e a s e h e l p w i t h w o r k i n g s .$
	Answered by sou1618 last updated on 17/Jul/16
	$x^2 =16^x →(∗) x^2 =(4^x )^2 ∣x∣=4^x (∵4^x >0) set f(x)=4^x −∣x∣ when x>=0 f ′(x)=(log_e 4)×4^x −1>0 f(0)=1 ∴f(0)>0 ⇒(∗)has no solution when x<0 f ′(x)=(log_e 4)×4^x +1>0 f(0)=1 lim_(x→−∞) f(x)=−∞ ⇒(∗)has only one solution −x=4^x (−x)^(1/x) =4 4= 2^2 =((1/2))^(−2) { ((−x=1/2)),(((1/x)=−2)) :} x=−(1/2)$
	$x^{2} = 16^{x} \to ()$ $x^{2} = {(4^{x})}^{2}$ $∣ x ∣= 4^{x} (∵ 4^{x} > 0)$ $set$ $f (x) = 4^{x} - ∣ x ∣$ $when x >= 0$ $f^{'} (x) = ({l o g}_{e} 4) \times 4^{x} - 1 > 0$ $f (0) = 1$ $∴ f (0) > 0 \Rightarrow () has no solution$ $when x < 0$ $f^{'} (x) = ({l o g}_{e} 4) \times 4^{x} + 1 > 0$ $f (0) = 1$ $\lim_{x \to - \infty} f (x) = - \infty$ $\Rightarrow (*) has only one solution$ $- x = 4^{x}$ ${(- x)}^{\frac{1}{x}} = 4$ $4 = 2^{2} = {(\frac{1}{2})}^{- 2}$ ${\begin{cases} - x = 1 / 2 \\ \frac{1}{x} = - 2 \end{cases}$ $x = - \frac{1}{2}$
	Commented by Tawakalitu. last updated on 17/Jul/16

	$T h a n k s s o m u c h .$
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