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?" ?