I know I will be ridiculed mercilessly, but my little brother's having trouble with his algebra and I don't remember anything from calculus.
Could any of you geniuses tell the answer (and how to get there) for:
3/(9^x)= 243
(2^x)(2^(x+1)) = 1/16
(4^x)^(x+2) = 64
God this is humiliating.
haha, yea, sure it's your little brother who's having problems
use logorithms...first one you need base 9, second one you need base 2, third one you need base 4...hope that helps
Solve for x:
3/(9x)=243
multiply by 3 on both sides, cancels out the 3 on the left:
9x = 729
divide by 9 on both sides, cancells out the 9 on the left:
x = 81
Plug 81 in x in the original equation to check.
That should get you started. Do the opposite to cancel out whatever you want to get ride of. Get the x on one side of the equation.
1) multiply by 9^x : 3 = 243(9^x)
divide by 243 : 3/243 = 9^x
take logarithms : ln (3/243) = ln (9^x)
use ln (a/b) = ln a - ln b, and ln (a^b) = b ln a
ln 3 - ln 243 = x ln 9
so x = (ln 3 - ln 243)/ln 9 = -2 (use ln button on your calculator!)
These are all contrived problems that don't require logarithms to solve them. Basic properties of exponents are sufficient:
3/9^x = 243
9^(-x) = 81
9^(-x) = 9^2
x = -2
====================
(2^x)(2^(x+1)=1/16
2^(2x+1) = 2^-4
2x+1 = -4
x = -5/2
====================
(4^x)^(x+2) = 64
(2^2x)^(x+2) = 2^6
2^(2x^2 + 4x) = 2^6
2x^2 + 4x = 6
2x^2 + 4x - 6 = 0
x^2 + 2x - 3 = 0
(x + 3)(x - 1) - 0
x = -3 or x = 1
(2^x)(2^(x+1)) = 1/16
(a^x)(a^y) = a^(x+y)
so (2^x)(2^(x+1)) = 2^(2x+1)
use logarithms again:
(2x+1) ln 2 = ln 1 - ln 16
(note ln 1 = 0):
re-arrange: x = - [(ln 16 / ln 2) - 1]/2 = -2.5
Alg II wrote:
These are all contrived problems that don't require logarithms to solve them. Basic properties of exponents are sufficient
Good think batman!
Yeah it's been too long since I've used this stuff (that's a tip to anyone out there wanting to be at math - practice lots of examples, so you know how to spot stuff to save you trouble). I suppose it was a valid method... anyway just trying to help a kid out.
I think now the OP should post a joke or something for our trouble!
You really suck
Well, at least give your age and total assets
make sure you check for extraneous roots
plug roots back into the original equations, see if they check out
Hey, Complete Idiot, at least you know how to spell "ridiculous" correctly. You're not a complete idiot.
Obviously your'e not Asian.
If you don't introduce even powers and don't use logs, you don't have to worry about extraneous solutions...
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