Thread: Exponential equations questions

Hi
i have tried to solve these equations but still cannot get the correct answer:
The following is what i have done:

make





then:







has no solution but answer says it is does.









P.S

Originally Posted by Paymemoney

has no solution but answer says it is does.

Given the arithmetic is correct

Originally Posted by Paymemoney

Hi
i have tried to solve these equations but still cannot get the correct answer:
The following is what i have done:

make





then:







has no solution but answer says it is does.









P.S

Your substitution should be...


Not

" 2y^2-20y+64=0 " is incorrect.

2^(2x)-20(2^x)=-64
" Let 2^x be y " is the right direction, but-

Your mistake: 2^(2x) = 2(2^x)^2.

2^(2x) = (2^x)(2^x) = 2^(x+x) = 2^(2x) = (y)(y) = y^2 ----( not 2y^2 )

As for the rest of the steps, I believe you know them already

Originally Posted by Paymemoney

Yikes! What are you doing? Keep it simple:

2^2(2^x) - 20(2^x) = -64
4(2^x) - 20(2^x) = -64
-16(2^x) = -64
2^x = -64 / -16
2^x = 4
x = 2

lols, Wilmer.

2^2(2^x) = 2^(x+2) is not the same as 2^(2x) = (2^x)(2^x) = 2^(x+x) = 2^(2x)

Originally Posted by werepurple

" 2y^2-20y+64=0 " is incorrect.

2^(2x)-20(2^x)=-64
" Let 2^x be y " is the right direction, but-

Your mistake: 2^(2x) = 2(2^x)^2.

2^(2x) = (2^x)(2^x) = 2^(x+x) = 2^(2x) = (y)(y) = y^2 ----( not 2y^2 )

As for the rest of the steps, I believe you know them already

oh oops, i think i'll tired it again and i should be able to get it.

lol, i didn't ever need to complete the square and i answer my own question.
i just made it:

which will therefore get me x=4 and x=2.

oh well, thanks for spotting the mistake

Originally Posted by werepurple

lols, Wilmer.
2^2(2^x) = 2^(x+2) is not the same as 2^(2x) = (2^x)(2^x) = 2^(x+x) = 2^(2x)

right...BUT tell me why I got the correct answer

Sheer coincidence, Wilmer. Haha, and that is why there are intersections of lines, curves, and etc; values coincide. ^^