# Math Help - Logarithms and surds

1. ## Logarithms and surds

To be exact, I have a few questions on logarithms.. The questions are listed as follow. It would be great if you could show necessary workings and solution.

Question 1
Prove that the equation has a real root from the root x=2. Find this root.

2^(2x) + 64[2^(-2)] =32

Question 2
Given that 2y = a^x + a^(-x) where a>1 and x>0
Prove that a^x = y + (y^2 -1)^1/2

Similarly if 2z = a^(3x) = a^(-3x) prove that z=4y^3 -3y

Question 3
Solve the equation
5/(x+1)^1/2 -3 (x+1)^1/2 + 2 = 0

2. 2.

$2y = a^x + a^(-x)$

$2y = a^x + \frac{1}{a^x}$

Multiply by a^x throughout:

$2y(a^x) = a^x(a^x) + 1$

$2y(a^x) = a^{2x} + 1$

$0 = a^{2x} - 2y(a^x) + 1$

Therefore, completing the square;

$0 = (a^x - y)^2 - y^2 + 1$

Rearrange for a^x;

$y^2 - 1 = (a^x - y)^2$

$(a^x - y) = (y^2 - 1)^{\frac12}$

$a^x = y + (y^2 - 1)^{\frac12}$

Ok, that's as far as I'm sure for this part. For the other one, I'm not quite getting something.

If z = a^3x = a^-3x;

a^x = z^1/3 = z^-1/3.

Taking a^x = z^1/3;

$z^{1/3} = y + (y^2 - 1)^{\frac12}$

$z = (y + (y^2 - 1)^{\frac12})^3$

Expanding, I get:

$z = 4y^3 -3y + (y^2-1)^{\frac12}(2y^2-1)$

Maybe it's because I didn't take a^x = z^-1/3, but if I do so, I'll get infinite terms, and I don't know how to deal with that.

3. $\frac{5}{(x+1)^{\frac12}} - 3(x+1)^{\frac12} + 2 = 0$

Is that it? If so, multiply throughout by (x+1)^1/2;

$5 - 3(x+1) + 2(x+1)^{\frac12} = 0$

You can make use of a substitution to help you solve here. Let (x+1)^1/2 = y;

$5 - 3y^2 + 2y = 0$

$3y^2 - 2y - 5 =0$

$(3y - 5)(y + 1)=0$

So, 3y -5 = 0,

y = 5/3

$(x+1)^{\frac12} = \frac53$

and y +1 = 0

y = -1

$(x+1)^{\frac12} = -1$

You can solve those two now.

For the first part, I prefer not answering because I don't quite understand what the question is asking for. I can solve for x, and get x = 2, but not show that it's a real root, sorry

3. Hello everyone

I'm sure the second part of number 2 should read:
$2z=a^{3x}+a^{-3x}$
So we can say:
$2z = (a^x)^3 + (a^{-x})^3$
$= (a^x+a^{-x})^3 - 3(a^x)^2(a^{-x})-3(a^x)(a^{-x})^2$, using $(p+q)^3 = p^3+3p^2q+3pq^2+q^3$

$= (a^x+a^{-x})^3 - 3\Big((a^x)+(a^{-x})\Big)$

$=(2y)^3-3\cdot2y$

$\Rightarrow z = 4y^3-3y$