Question

Please explain how to solve:

If x1 and x2 are solutions of the equation

2^4x^2 = 1/4^4x+1

compute the value of x1-x2

Problem 2:

Find y when
2^x = 16^y, 16^x = 8^y+5

Answer 1

Is this the equation for the first part? It's difficult to infer what you mean if it's typed out like that. I'd suggest in the future typing anything complex out using the equation editor in a comment reply.
\[2^{4x^2} = \frac{1}{4^{4x}}+1\]

Answer 2

I'm sorry I didn't know of the equation editor. This is the equation:

\[2^{4x ^{2}}=1/(4^{4x+1})\]

Answer 3

First of all I'd bring the denominator on the right up to the left hand side. Then I'd consider rewriting the 4 (the one that isn't inside an exponent) and see if that helps at all (looks like it will!).

Answer 4

Thank you for the reply, these are the steps I did. I'm stuck and really have no idea how to solve it.

\[2^{4x ^{2}}(4^{4x+1})=1\]
\[8^{4x ^{2}+8x+2}=1\]
\[\log _{8}(1) = 4x ^{2} + 8x + 2\]

Answer 5

i think there is a mistake here

Answer 6

\[4^{4x+1}=(2^2)^{4x+1}=2^{8x+2}\]

Answer 7

then
\[2^{4x^2}\times 2^{8x+2}=2^{4x^2+8x+2}\]

Answer 8

this tells you
\[4x^2+8x+2=0\] or
\[2x^2+4x+1=0\] which you can solve using the quadratic formula ,

Answer 9

Thank you so much, I now see what I was doing wrong.

Answer 10

oh wait, now i see the problem is something else entirely

Answer 11

But if I find x won't I be able to just plug in and solve?

Answer 12

\[2^{4x ^{2}}=\frac{1}{4^{4x+1}}\]

Answer 13

is that the problem?

Answer 14

Yes

Answer 15

If x1 and x2 are solutions compute the value of x1-x2

Answer 16

if you write them with the same base of 2, you get
\[2^{4x^2}=(2^{-2})^{4x+1}\]
\[2^{4x^2}=2^{-8x-2}\]

Answer 17

now you have to solve
\[4x^2=-8x-2\] oh i see, it is the same...

Answer 18

solve
\[2x^2+4x+1=0\]

Answer 19

The logarithm of 1 in any base is 0. =))

Answer 20

Thank you both, I got the answer I was looking for.