AQA Pure Core 2 AS Maths Exam 25th May

First of all lets make all logs have the same base.
$\displaystyle \log_{100}[(2x+6)^2]=\frac{\log_{10}[(2x+6)^2]}{\log_{10}100} =\frac{1}{2}\log_{10}[(2x+6)^2]$ then using the power rule you get $\log_{10}(2x+6)$. As all logs in the equation have base 10 I'm not going to bother putting the base in any more.

$\displaystyle \log[(x-2)(x+2)]+\log[2(x+3)]=\log(x+2)+\log(x+3)$
$\Rightarrow \log(x-2)+\log(x+2)+\log 2 +\log(x+3)=\log(x+2) + \log(x+3)$. Notice 2 of the terms cancel on both sides of the equation.
This leaves
$\displaystyle \log(x-2)+\log(2) = 0 \Rightarrow \log(2x-4)=0$
this obviously means that
$\displaystyle 2x-4=10^0=1 \Rightarrow \boxed{ x=5/2 }$

WILL I BE GIVEN TRAPEZIUM RULE FORMULA AND other formulas like sum to infinity UN e.t.c.

How much revision have you guys done?
Done all the past papers and looked at some of the Solomon ones but always mess up in the questions with lots of marks.

Thanks my G. Do u think we'll get a log question that hard

what paper is that question

I have never learnt change the bases log rule. If it comes up I'm going to drink bleach

I have never learnt change the bases log rule. If it comes up I'm going to drink bleach

Thought that exam was quite nice. How did everyone else find it?