# A-Math: Spot the Mistake in this Logarithm Equations

I gave my Secondary 4 students 1 revision question per week. The recent one I marked was on the topic of Logarithms (the fresh-killer topic for many Secondary 3).

Did you spot the error? (It's a give-away if you observe clearly)

I look forward in hearing from you.

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### Ai Ling Ong

Hi, I'm Ai Ling Ong. I enjoy coaching students who have challenges with understanding and scoring in 'O' Level A-Maths and E-Maths. I develop Math strategies, sometimes ridiculous ideas to help students in understanding abstract concepts the fast and memorable way. I write this blog to share with you the stuff I teach in my class, the common mistakes my students made, the 'way' to think, analyze... If you have found this blog post useful, please share it with your friends. I will really appreciate it! :)

### 6 Responses to A-Math: Spot the Mistake in this Logarithm Equations

1. nirosh tharaka says:

there is a wrong calculation in divition of logarithm

2. nil says:

(log2(base2)8)divide by(log2(base2)x)
is not equal to (log2(base2)8/x)

(log2(base2)8)-(log2(base2)x)=log2(base2)8/x)

3. AMath Student says:

Let log2 x be y.
2y = 3(log2 2 over log2 x) + 1
2y - 1 = 3 ( 1 over y )
2y^2 - y = 3
2y^2 - y - 3 = 0
( 2y - 3 )( y + 1 ) = 0
2y = 3 , y = -1
y = 3/2

If there is a faster method, i would like to know ;)