Need Math help!!
 

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?

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

give me a sec.

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..

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

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

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

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)

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

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

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