Question

how would you solve for \[6^{2x} - 2(6^x)-15=0\]

Answer 1

do i evaluate it as 6^(2x) - 12^x - 15 = 0 first?

Answer 2

this looks like a disguised quadratic equation

Answer 3

If you let \(\large u = 6^x\)
then \(\large u^2 = 6^{2x}\)
so \(\large u^2 - 2u - 15 = 0\)

Answer 4

okay so from there... i factor it out right?

Answer 5

right.

Answer 6

Oh... wow. okay so for this case i keep the 2 out. and don't multiply it with 6 then?

Answer 7

right - order of operations - 2 * 6^x is not equal to 12^x

Answer 8

for example if x is 2 then
2*6^2 = 2*36 = 72
but
12^2 = 144

Answer 9

wait so i got (u+5)(u-3) when i factored right..
then 2(6) ... huh??? :S omg i got lost again..

Answer 10

lol...I was answering your side question about 2*6^x verses 12^x with an example

Answer 11

you should get (u-5)(u+3) when you factor

Answer 12

oh lol ok oh and ye so i then reject the negative and get 3.
6^x = 3
ln(6)^x=ln(3)
xln(6)=ln(3)
... uhh... ? what now?
... wait isn't it (u+5)(u-3)? not (u-5)(u+3)

Answer 13

yep, the signs...

Answer 14

middle term is negative, so larger factor is negative

Answer 15

wait woah. i just confused myself... -.-

Answer 16

sorry! nvm what i said. ooook. soo then
it would be x = ln(3)/ln(6)

Answer 17

almost...but.... when you factor:
\[\large u^2 - 2u - 15 = (u-5)(u+3)\]
so u = 5 and u = -3
reject the negative answer and then
\[\large 6^x = 5\]
\[\large x = \frac{\ln5}{\ln6}\]

Answer 18

iofgiohda omg ! .. i forgot to switch the signs. ok so i have another q. and it is \[4^{2x} + 9(4^x) + 14=0\]

and i got two negative roots... would that mean that x = 0 and i don't do anything else to it?

Answer 19

looks like no solution then

Answer 20

okay thank you again. im sorry if i might be really bad at this... the notes given by the teacher are really poorly written.

Answer 21

you're really not bad at this at all.

Answer 22

oh well thanks haha