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

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- Dec 6th 2009, 08:43 PMPaymemoneyExponential 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 - Dec 6th 2009, 08:49 PMpickslides
- Dec 6th 2009, 08:50 PMVonNemo19
- Dec 6th 2009, 08:56 PMwerepurple
" 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 :D - Dec 6th 2009, 08:56 PMWilmer
- Dec 6th 2009, 09:00 PMwerepurple
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) - Dec 6th 2009, 09:01 PMPaymemoney
- Dec 6th 2009, 09:06 PMPaymemoney
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 - Dec 6th 2009, 09:08 PMWilmer
- Dec 6th 2009, 09:10 PMwerepurple
Sheer coincidence, Wilmer. Haha, and that is why there are intersections of lines, curves, and etc; values coincide. ^^