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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
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are u saying that u can't use natural log?
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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!!
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cool. u figured it out?
were any of use even close? |
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yup, that worked, thank you for your help! In other forums, I was riddiculed, but here, people helped, so once again, thank you very much
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...I think I missed that class....:rolleyes:
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