# Exponent Problem

• Nov 21st 2011, 12:00 AM
BobBali
Exponent Problem
Hi All & Good-day,

The problem states, Show that:

$(\frac{2}{x} + 4)^2 = \frac{1 + 2^x+3 + 2^2x+4}{2^2x }$

* Above Denimonator should be 2 power 2x (latex trouble)

$(2^{-x + 4})(2^{-x + 4})$

$2^{-2x + 4}(2^{-x}) + 4(2^{-x}) + 16$

$2^{-2x} + 2^{-x} (4+4) + 16$

$2^{-2x} + 2^{-x} (2^2 + 2^2) + 2^4$

$2^{-2x} + 2^{-x+2} + 2^{-x+2} + 2^4$

After this i am at a loss... But i do know that if i take the above first term's
reciprocal > $\frac{1}{2^2x }$

I now have the denominator in the answer, but how to distinguish the 2nd and 3rd terms in the answer...?
• Nov 21st 2011, 03:28 AM
Amer
Re: Exponent Problem
Quote:

Originally Posted by BobBali
Hi All & Good-day,

The problem states, Show that:

$(\frac{2}{x} + 4)^2 = \frac{1 + 2^x+3 + 2^2x+4}{2^2x }$

* Above Denimonator should be 2 power 2x (latex trouble)

$(2^{-x + 4})(2^{-x + 4})$

$2^{-2x + 4}(2^{-x}) + 4(2^{-x}) + 16$

$2^{-2x} + 2^{-x} (4+4) + 16$

$2^{-2x} + 2^{-x} (2^2 + 2^2) + 2^4$

$2^{-2x} + 2^{-x+2} + 2^{-x+2} + 2^4$

After this i am at a loss... But i do know that if i take the above first term's
reciprocal > $\frac{1}{2^2x }$

I now have the denominator in the answer, but how to distinguish the 2nd and 3rd terms in the answer...?

for the first one is not clear in the left hand side there is not $2^x$ how in the right side there is ??
for the second one

$(2^{-x+4} ) (2^{-x+4}) = 2^{-x+4-x+4} = 2^{-2x+8} = 2^{-2x} \cdot 2^8$
• Nov 21st 2011, 04:33 AM
HallsofIvy
Re: Exponent Problem
Quote:

Originally Posted by BobBali
Hi All & Good-day,

The problem states, Show that:

$(\frac{2}{x} + 4)^2 = \frac{1 + 2^x+3 + 2^2x+4}{2^2x }$

* Above Denimonator should be 2 power 2x (latex trouble)

This is obviously NOT true. If x= 1, the left side is $6^2= 36$ while the left side is $\frac{14}{4}= \frac{7}{2}$

To put more than one character in an exponent use ^{ }. For example $2^{2x}$ requiires 2^{2x}.
• Nov 21st 2011, 10:48 AM
BobBali
Re: Exponent Problem
Sorry the RHS of the equation is supposed to look like this>
$1 + 2^{x+3} + 2^{2x+4} \div 2^{2x}$
• Nov 21st 2011, 08:03 PM
Amer
Re: Exponent Problem
Quote:

Originally Posted by BobBali
Sorry the RHS of the equation is supposed to look like this>
$1 + 2^{x+3} + 2^{2x+4} \div 2^{2x}$

what about the left hand side, use brackets
• Nov 21st 2011, 11:29 PM
BobBali
Re: Exponent Problem
The LHS is as was shown originally> $(2^{-x} + 4) (2^{-x} + 4)$
• Nov 22nd 2011, 04:10 AM
Soroban
Re: Exponent Problem
Hello, BobBali!

Quote:

$\text{Show that: }\:\left(2^{-x} + 4\right)^2 \:=\: \frac{1 + 2^{x+3} + 2^{2x+4}}{2^{2x} }$

$\left(2^{-x} + 4\right)^2 \;=\;\left(\frac{1}{2^x} + 2^2\right)^2 \;=\;\left(\frac{1 + 2^2\!\cdot\!2^x}{2^x}\right)^2 \;=\;\left(\frac{1+2^{x+2}}{2^x}\right)^2$

. . . . . . . . $=\;\frac{1 + 2\!\cdot\!2^{x+2} + 2^{2(x+2)}}{(2^x)^2} \;=\;\frac{1 + 2^{x+3} + 2^{2x+4}}{2^{2x}}$