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

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Homework Help Overview

The problem involves solving the equation 3(6x-1) = 1 + 4/6x, with a specific requirement to express the solution in the form x = a - log6b, where a and b are integers. The subject area pertains to logarithmic functions and algebraic manipulation.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to manipulate the equation by substituting 6x with y and forming a quadratic equation. Some participants question the steps taken and suggest verifying the manipulation of terms. Others raise the need to take logarithms of both sides and inquire about solving equations of the form Ax = B.

Discussion Status

The discussion is ongoing, with participants providing guidance on taking logarithms and exploring the implications of the original equation. There is an exchange of ideas regarding the steps taken and the interpretation of the equation, but no consensus has been reached on the next steps.

Contextual Notes

Participants express feelings of confusion related to the complexity of the problem, particularly in the context of their studies in IB HL Maths. There is an acknowledgment of the challenges faced in understanding logarithmic functions.

thornluke
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Homework Statement


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


Homework Equations





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? :confused:
 
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thornluke said:
3(6x)(6-1)(6x) = 6x + 4
Let 6x be y,
y = 1 ± 3
3x = 4
xlog63 = log64

Read the bolded statements carefully. :wink:
 
Infinitum said:
Read the bolded statements carefully. :wink:

OH NO! What is IB HL Maths doing to me :eek:
Thanks! :approve:
 
thornluke said:
OH NO! What is IB HL Maths doing to me :eek:
Thanks! :approve:

Its normal, after lots of studying!
Take short breaks, now and then :biggrin:
 
Isn't it 6x = 4?
 
Yes, now take the logarithm, with respect to any base, of both sides.

By the way, going back to your first formula, [itex]x log_6(3)= log_6(4)[/itex], how would you solve any equation of the form Ax= B for x?
 
HallsofIvy said:
Yes, now take the logarithm, with respect to any base, of both sides.

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

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

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