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i really need ur help. (algebra)?
its our exam, and i really can't answer some questions in our reviewer.
clarification please.
is 4x^-1 same as 1/4x or is it 4/x?
find the value of x:
1.) log(x^1/2) = (logx)^1/2
log^2x = logx(then? i don't know whats next..) :'c
find the value of x and y
2.) x^logx = (10^4)x (i really don't know what to do in this eq'n)
3/x + 2/y = 3/2
(elimination or substitution will do)
3.) 2/x - 4/y = -1/3
3x-2y-3z = 2
no z in the first eq'n, what to do?
______________________________________________
Given: f(x) = (1-x^2/3)^3/2
Find: f[f(x)]
pls i need ur help. ASAP. and f u could give some exercises and its answers but not the sol'n. i'm troubled in solving logarithms.. some tips pls in solving logarithms and word problems, i'm having a hard time solving this and still i can't get the ans.
kindly explain ur answer too..
tnx in advance..:)
1 Answer
- Anonymous1 decade agoFavorite Answer
4(x)^-1 = 4/x
(4x)^-1 = 1/4x
It depends on what the power applies to. If they just put the power next to x without any brackets to include 4, its 4/x. If they used brackets to include number 4, its 1/4x.
1. log(x^1/2) = (logx)^1/2 [I assume log means base 10]
1/2 lg x = (lgx)^1/2
1/4 (lgx)^2 = lgx
1/4 (lgx)^2 - lgx = 0
(lgx)^2 - 4lgx = 0
To make it easier to see, let y = lgx
y^2 - 4y = 0
y(y - 4) = 0
y = 0 or y = 4
lgx = 0 or lgx = 4
x = 1 or x = 10000
2. x^(logx) = (10^4)x
[This part involves changing y=a^x => log basea y = x]
lgx = log base(x) [(10^4)x]
lgx = log base(x) {10^4) + log base(x) x
lgx = 4 log base(x) 10 + 1
lgx = 4 / lgx + 1
(lgx)^2 = 4 + lgx
(lgx)^2 - lgx - 4 = 0
To make it easier to see, let y = lgx
y^2 - y - 4 = 0
y = [ 1 +/- sqrt(17) ] / 2
lgx = [ 1 + sqrt(17) ] / 2 or lgx = [ 1 - sqrt(17) ] / 2
x = 364(3s.f.) or 0.0274(3s.f.)
3/x + 2/y = 3/2 [I assume its the same x values as the previous?]
2/y = 3/2 - 3/x
2/y = 3x/2x - 6/2x = (3x-6) / 2x
y/2 = 2x / (3x-6)
y = 4x / (3x - 6)
y = 1.34(3s.f.) or -0.0186(3 sf.)
Not sure on the last 2 though, sorry.