# Math Help - Exponent Problem

1. ## 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...?

2. ## Re: Exponent Problem

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

$(2^{-x+4} ) (2^{-x+4}) = 2^{-x+4-x+4} = 2^{-2x+8} = 2^{-2x} \cdot 2^8$

3. ## Re: Exponent Problem

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}.

4. ## Re: Exponent Problem

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

5. ## Re: Exponent Problem

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

6. ## Re: Exponent Problem

The LHS is as was shown originally> $(2^{-x} + 4) (2^{-x} + 4)$

7. ## Re: Exponent Problem

Hello, BobBali!

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

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

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