1. ## Log questions

This is for IB standard level math which is basically a review of everything through calculus and I'm having trouble remembering how to do some of this stuff. I have a test tomorrow so if anyone could help me out, it would be greatly appreciated.

1. Solve for X and Y using a system of equations
4^X x 8^Y = 2
3^3X x 9^2Y = 27^(2/3)

2. If log(base a)2 = X and log(base a)5 = Y, find...
a. log(base 2)5
b. log(base a)20

3. Solve the system:
Y = 2log(base 3)X
Y+1 = log(base 3)9X

4. Use the first three terms in the expansion of (1+X)^5 to find an approximate value for 1.06^5

5. If log(base b)X = 5 and log(base b)sqrtY = 12, find...
a. log(base b)b^Y

I know it's not very advanced stuff but I'm having trouble with this class and I'd like to keep my GPA as high as I can until college apps are finished.

Thanks.

2. Originally Posted by Springs
This is for IB standard level math which is basically a review of everything through calculus and I'm having trouble remembering how to do some of this stuff. I have a test tomorrow so if anyone could help me out, it would be greatly appreciated.

1. Solve for X and Y using a system of equations
4^X x 8^Y = 2
3^3X x 9^2Y = 27^(2/3)

Thanks.
$\displaystyle 4^x \cdot 8^y = 2 \Longrightarrow 2^{2x} \cdot 2^{3y}=2^1 \Longrightarrow \boxed{2x+3y=1}$

$\displaystyle 3^{3x} \cdot 9^{2y}=27^{\frac{2}{3}} \Longrightarrow 3^{3x} \cdot 3^{4y}=3^2 \Longrightarrow \boxed{3x+4y=2}$

Now solve for x and y by substitution, elimination, or matrices, whichever method you know best.

3. 1. Solve for X and Y using a system of equations
4^X x 8^Y = 2
3^3X x 9^2Y = 27^(2/3)
hints ...
$\displaystyle 4^x = (2^2)^x$
$\displaystyle 8^y = (2^3)^y$

$\displaystyle 9^{2y} = (3^2)^{2y}$
$\displaystyle 27^{\frac{2}{3}} = (3^3)^{\frac{2}{3}}$

2. If log(base a)2 = X and log(base a)5 = Y, find...
a. log(base 2)5
b. log(base a)20
hint for part (a) ... $\displaystyle \log_b(c) = \frac{\log_a(c)}{\log_a(b)}$

hint for part (b) ... $\displaystyle 20 = 2^2 \cdot 5$ and
$\displaystyle \log(b^2c) = \log(b^2) + \log(c)$

3. Solve the system:
Y = 2log(base 3)X
Y+1 = log(base 3)9X
hint ... $\displaystyle \log_3(9x) = \log_3(9) + \log_3(x)$

4. Use the first three terms in the expansion of (1+X)^5 to find an approximate value for 1.06^5
hint ...
$\displaystyle (a+b)^5 = a^5 + 5a^4b + 10a^3b^2 + 10a^2b^3 + 5ab^4 + b^5$

5. If log(base b)X = 5 and log(base b)sqrtY = 12, find...
a. log(base b)b^Y
hints ...
$\displaystyle \log_b{\sqrt{y}} = \frac{1}{2}\log_b{y}$
$\displaystyle \log_b{b^y} = y$