# Logarithmic Functions

1. May 11, 2012

### thornluke

1. The problem statement, all variables and given/known data
Solve 3(6x-1) = 1 + 4/6x giving your answer in the form x = a - log6b where a,b are integers (Z)

2. Relevant equations

3. The attempt at a solution
3(6x)(6-1)(6x) = 6x + 4
Let 6x be y,
(1/2)y2 - y - 4 = 0
y = 1 ± √(1+8) = 1 ± 3
3x = 4
xlog63 = log64

What should I do next?

2. May 12, 2012

### Infinitum

3. May 12, 2012

### thornluke

OH NO! What is IB HL Maths doing to me
Thanks!

4. May 12, 2012

### Infinitum

Its normal, after lots of studying!!
Take short breaks, now and then

5. May 12, 2012

### Staff: Mentor

Isn't it 6x = 4?

6. May 13, 2012

### HallsofIvy

Staff Emeritus
Yes, now take the logarithm, with respect to any base, of both sides.

By the way, going back to your first formula, $x log_6(3)= log_6(4)$, how would you solve any equation of the form Ax= B for x?

7. May 13, 2012

### thornluke

What do you mean by "form Ax= B for x?" ?