Need Math help.

Is this like A Level maths? Ill try work it out for you.

A| Y = 0 (I think)

 

To help you with A, log base x of x^2 = 2.

 

I'll be doing Part A in a little bit I'm hoping.

There's a rule that I re-learned earlier this year that I'm hoping will assist in solving it.

Part B

In the first part of the equation, you can square both sides by x, which will cancel out the x. This will leave you with:
z = y^2

Now that this is established, start plugging in numbers with perfect squares.

F(4): 4 = 2^2

F(4): 2^4 = 2 * 4^x
F(4): 16 = 2 * 4^x
F(4): 8 = 4^x
F(4): x = 1.5

1.5 + 2 + 4 = 7.5

z =/= 4

-------------------------------

F(9): 9 = 3^2

F(9): 2^9 = 2 * 4^x
F(9): 512 = 2 * 4^x
F(9): 256 = 4^x
F(9): x = 4

4 + 3 + 9 = 16

^ Answer ^

 

Part A :S

(log[3]x)(log[x]2x)(log[2x]y)=log[x]x^2
((log[x]x)/log[x]3))X(log[x]2x)X(log[x]y)/(log[x]2x) = 2log[x]x
(1/(log[x]3)X(log[x]y) = 2
-(log[x]3)X(log[x]y) = 2
Hmm stuck :S

umm for part B I did this

2^x = y^2x
xlog2 = 2xlogy
2log2 = logy

2^z = 2.4^x
zlog2 = xlog2.4
z = x(log2.4/log2)

x + y + z = 16
x + y + x(log2.4/log2) = 16
2x + y = 16/(log2.4/log2) 16/(log2.4/log2) = A from now on
log2x + logy = logA
log2x = logA
2x = 10^logA
2x = 3.165
x = 1.5825

2x + y = A
A-3.165 = y
y = 9.5

x + y + z = 16
16-9.5-1.5825 = z
z=4.914

Idk If thats wrong or not btw I just gave it my best shot

 

Just wondering as to how squaring both sides by x will cancel out the x on the left side, if someone could elaborate, it'd be great.

 

First one equals 9. You just expand it out and all the terms with x cancel out. Then you can find y.

If you learned. Acually he worded it wrong. It's not squaring by x, its rooting by x. Squaring by x will just make the thing more complicated. Rooting by x also means raising it to the 1/x power. And when you exponent and exponent you multiply them. So x * 1/x = 1 canceling out the x.

EDIT: Opps i meant 9.

 

this looks hard

 

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