# Math Help - (UK) A-level Maths H/W.

1. ## (UK) A-level Maths H/W.

Hi, i used this forum for my GCSE mathematics, and I managed to get an A in the end! Firstly, if any of you remember my panicked questions, i would like to thank you for all the help given as similar questions came up in the test requiring that method to solve them! I moved to higher tier 4 weeks before i broke up from school, so had to ask friends and you helpful people for help, but thankfully it was worth it and i got the A.. so thank you all who helped!

Anyway.. I have moved on. I am now doing A2 level maths along with biology, chemistry, english language (and a spanish course for beginners). My teacher has given me some terribly hard questions and i was wondering if anyone could get me going on these.. the whole class thinks they are hard but i need to try and answer some! So hear goes..

1a) Given that 2^a = 8^b, express a in terms of b.
1b) State the value of x for which 9^x = 3.
2) If x^p = (√x)^3, x^q = 1/x^2 and x^r=x^p/x^q
a) Find the value of x
b) Evaluate x^r when x = 4
3a) Express (1+√2)^2 in the form a+b√2, where a and b are integers to be found.
3b) By substituting x = y^1/2, or otherwise, solve the equation 6y^1/2 = y+9

There are more questions.. many more but to get me started some help on those would be great

2. Originally Posted by Danielisew
1a) Given that 2^a = 8^b, express a in terms of b.
Do you remember the properties of logarithms?
$2^a = 8^b$

$2^a = (2^3)^b$

$2^a = 2^{3b}$

$log_2(2^a) = log_2(2^{3b})$

$a = 3b$

Originally Posted by Danielisew
1b) State the value of x for which 9^x = 3.
$9^x = 3$

$(3^2)^x = 3^1$

$3^{2x} = 3^1$

$log_3(3^{2x}) = log_3(3^1)$

$2x = 1$
etc.

-Dan

3. Originally Posted by Danielisew
2) If x^p = (√x)^3, x^q = 1/x^2 and x^r=x^p/x^q
a) Find the value of x
b) Evaluate x^r when x = 4
Try to do these the same way I did the last one. You will need to recall that
$\sqrt{x} = x^{1/2}$

-Dan

4. Originally Posted by Danielisew
3a) Express (1+√2)^2 in the form a+b√2, where a and b are integers to be found.
You should be able to do this one!
$(1 + \sqrt{2} )^2 = (1 + \sqrt{2} )(1 + \sqrt{2} )$

Now FOIL it out:
$= 1 \cdot 1 + 2 \sqrt{2} + \sqrt{2} \cdot \sqrt{2} =$?

Originally Posted by Danielisew
3b) By substituting x = y^1/2, or otherwise, solve the equation 6y^1/2 = y+9
You want to rewrite this in the form of a quadratic equation:
$6y^{1/2} = y + 9$

$y - 6y^{1/2} + 9 = 0$

Now let $z = y^{1/2}$, so $z^2 = y$, giving:
$z^2 - 6z + 9 = 0$

Factor this and solve it. Don't forget to take the square root of z at the end to get your y.

-Dan

5. Hello, Danielisew!

For most of these, you're expected to know that:

. . If $b^x\:=\:b^y$, then $x = y.$

Given two equal exponentials with equal bases, their exponents are equal.

1a) Given that $2^a \,= \,8^b$, express $a$ in terms of $b.$
We have: . $2^a\:=\:(2^3)^b\quad\Rightarrow\quad 2^a\:=\:2^{3b}$

Therefore: . $\boxed{a \:=\:3b}$

1b) State the value of $x$ for which $9^x \,= \,3.$
We have: . $(3^2)^x\:=\:3\quad\Rightarrow\quad3^{2x} \:=\:3^1$

Therefore: . $2x \:=\:1\quad\Rightarrow\quad\boxed{ x \:=\:\frac{1}{2}}$

2) If $x^p = (\sqrt{x})^3,\;x^q = \frac{1}{x^2}\;\text{and }x^r = \frac{x^p}{x^q}$

a) Find the value of $x$ . . . . Did you mean ${\color{blue}r}$ ?
$x^p \:=\:(\sqrt{x})^3\quad\Rightarrow\quad x^p \:=\:x^{\frac{3}{2}}\quad\Rightarrow\quad p \:=\:\frac{3}{2}$

$x^q \:=\:\frac{1}{x^2}\quad\Rightarrow\quad x^q \:=\:x^{-2}\quad\Rightarrow\quad q \:=\:-2$

Then: . $x^r\:=\:\frac{x^{\frac{3}{2}}}{x^{-2}}\quad\Rightarrow\quad x^r\:=\:x^{\frac{7}{2}}\quad\Rightarrow\quad\boxed { r \:=\:\frac{7}{2}}$

b) Evaluate $x^r$ when $x = 4$
$\text{If }x = 4\!:\;\;x^r \;=\;4^{\frac{7}{2}} \;=\;\boxed{128}$