Question

idk if my brain is just turning off because it's late... but would someone please help me with solving these two equations?

\[4^x + 6(4^{-x}) =5\]
\[8(5^{2x}) + 8 (5^x) = 6\]

Answer 1

if you multiply the first one by 4^x
you get a quadratic in 4^x

Answer 2

Hint: for the first equation, a quadratic in 4^x

Answer 3

second one is a similar idea, but it's already 'cleaned up' for you

Answer 4

e u=4^x
solve u^2 -5u +6 =0 for u...
then use that in u=4^x to find x



(fixed a sign)

Answer 5

Let u be 4^x

\[u + 6u^{-1} = 5\]Multiply both sides by u.\[u(u + 6u^{-1}) = 5u\]\[u^2 + 6 = 5u\]\[u^2 - 5u + 6 = 0\] Use quadratic formula to solve for u.

Answer 6

@alexeis_nicole ?

Answer 7

hmm. sooo. when i solved using the quadratic formula i got u = 3, 2 @micahwood50

Answer 8

it's factorable actually, but yeah.

Answer 9

Correct. Since we now know that\[u = 4^x\]
Now plug in values then find the x.

Answer 10

so x = 64, or 16 ?

Answer 11

3=4^x , 2=4^x
ln3/ln4 =x, ln2/ln4 =x

Answer 12

oooh wait

Answer 13

whooops my bad

Answer 14

OR: \[\Large x = \log_4 3 \text{ and }\log_4 2 \text{ or } \frac{1}{2}\]

Answer 15

its on 0.79 and 0.5

Answer 16

Yeah. That's right.

Answer 17

Is this clear? Can you solve second problem?