Question

Whe does this happen with exponents?

Answer 1

i saw in a book that if,
(a+b)^n =x then (a-b)^n = 1/x

Why does this happen? Is this true?

Answer 2

3^2 = 9

0^2 = 0 not= 1/9

Answer 3

hmm ... lets try that again

Answer 4

(2+1)^2

(2-1)^2

yeah, thats still not working always

Answer 5

in general, it looks dubious
(a+b)^n =x then (a-b)^n = 1/x

if true then
(a+b)^n * (a-b)^n = x * 1/x
( (a+b)(a-b) )^n = 1
(a^2 -b^2)^n = 1

I can think of lots of a,b numbers where this is not true.

Answer 6

beaten to the punch.

Answer 7

perhaps you mean
\[ (a+b)^n = x \]
then
\[ (a+b)^{-n} = \frac{1}{(a+b)^n} = \frac{1}{x} \]

Answer 8

This was the question and i was just looking at it's solution? I still can't figure out how to solve this??

if \[(5+ 2\sqrt{6})^{x ^{2}-3} + (5- 2\sqrt{6})^{x ^{2}-3} =10 \] =10 then x is equal to?

Answer 9

@amistre64 @phi

Answer 10

this looks difficult.

Answer 11

Sorry, I had to run out for some errands, but from phi's manipulation, we get that
\[ (a+b)^n = \frac{1}{(a-b)^n} \] iff \[ a^2 - b^2 = 1 \]

In this case, \[ a = 5 \: \:\: b = 2\sqrt{6} \\ a^2 -b^2 = 25-24 = 1 \]

We can then use this property to help solve our equation.

\[(5+2\sqrt{6})^{x^2-3} +(5-2\sqrt{6})^{x^2-3} = (5+2\sqrt{6})^{x^2-3} +\frac{1}{(5+2\sqrt{6})^{x^2-3}} = 10 \\ \frac{(5+2\sqrt{6})^{(x^2-3)^2}+1}{(5+2\sqrt{6})^{x^2-3}} = 10 \to (5+2\sqrt{6})^{(x^2-3)^2} -10(5+2\sqrt{6})^{(x^2-3)}+1 = 0 \] Let \[ z = (5+2\sqrt{6})^{(x^2-3)} \] The eqn then becomes:
\[z^2 -10 z + 1 = 0 \\ z = \frac{10\pm\sqrt{96}}{2}\]
Substitute z back in \[ (5+2\sqrt{6})^{x^2-3} = \frac{10\pm\sqrt{96}}{2}\]

Solve with logarithms, 4 solutions exist.

Answer 12

nice insight from @kmeis002

notice if we simplify the right-hand side we get
\[ (5+2\sqrt{6})^{x^2-3} = 5± 2 \sqrt{6} \]
if we assume the positive root we find
\[ (5+2\sqrt{6})^{x^2-3} = (5+ 2 \sqrt{6})^1 \]
with equal bases, we can equate the exponents to find
\[ x= ± 2 \]

for the negative root
\[ (5+2\sqrt{6})^{x^2-3} = 5- 2 \sqrt{6} \\
(5+2\sqrt{6})^{x^2-3}(5+2\sqrt{6})= (5-2\sqrt{6})(5+2\sqrt{6}) \\
(5+2\sqrt{6})^{x^2-2} = 1
\]
from which we find x^2-2 must be 0 and
\[ x= ± \sqrt{2} \]