Need Math help!!

Chip · #1 03-16-2006, 05:56 PM

My son's alg 2 homework is something I can't figure out without logs.

Solve for x:

2^x * 2^(x+1)=(1/8)

for do I do this w/out logs?

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broknlgs24 · #2 03-16-2006, 06:02 PM

is this 2 to the x times 2 to the (x + 1) = 1/8?

give me a sec.

Chip · #3 03-16-2006, 06:06 PM

yes, it is 2 to the power x times 2 to the power x+1 equaling 1/8.

the answer with the graphing calc (ti89) comes out to -2, but how..

broknlgs24 · #4 03-16-2006, 06:14 PM

ok. when u multiply 2 numbers that have exponents,
add the exponents. so, it will be 2^2x+1 = 1/8

still working on it though

Chip · #5 03-16-2006, 06:17 PM

ok cool, thnx for the help.. havent doe it this way in a long time

broknlgs24 · #6 03-16-2006, 06:24 PM

are u saying that u can't use natural log?

broknlgs24 · #7 03-16-2006, 06:33 PM

2^2x+1 = (1/8)
2x+1(ln 2) = ln (1/8)
divide both sides by ln2. that leaves
2x+1 = (ln (1/8)/ln 2)
subtract 1 from each side that leaves
2x = ((ln (1/8)/ln 2) -1) divide both sides by 2 that leaves
x = (((ln (1/8)/ln 2) - 1)/2)

ogx5 · #8 03-16-2006, 07:11 PM

Let's see, it's been a few years:

2^x * 2^(x+1) = 1/8 = 2^(-3)
2^(2x+1) = 2^(-3)
2x+1 = -3
x = -1

x5GuyInLA · #9 03-16-2006, 07:15 PM

how about 2^x * 2^(x+1) = (1/8) = 2^-3
x + x + 1 = -3
x = -2

i know it's simplified, but I don't remember how to get rid of the 2 (I assume that's what natural log or inverse log does), but logic tells me that that works. Wow, it's been way too long.....i think i've gotten dumber with age

Chip · #10 03-16-2006, 07:27 PM

can use ln, natural logs, sry, thnx anyway, i got it w/your help!!

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