Master Logarithmic Functions: Solve Equations with Integers | 3(6x-1) = 1 + 4/6x

[thornluke]

Homework Statement

Solve 3(6^x-1) = 1 + 4/6^x giving your answer in the form x = a - log₆b where a,b are integers (Z)

Homework Equations

The Attempt at a Solution

3(6^x)(6^-1)(6^x) = 6^x + 4
Let 6^x be y,
(1/2)y² - y - 4 = 0
y = 1 ± √(1+8) = 1 ± 3
3^x = 4
xlog₆3 = log₆4

What should I do next?

 

[Infinitum]

3(6^x)(6^-1)(6^x) = 6^x + 4
Let 6^x be y,
y = 1 ± 3
3^x = 4
xlog₆3 = log₆4

Read the bolded statements carefully.

 

[thornluke]

Read the bolded statements carefully.

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

 

[Infinitum]

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

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

 

[NascentOxygen]

Isn't it 6^x = 4?

 

[HallsofIvy]

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?

 

[thornluke]

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?

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