There we discussed the need for extension of Exponents and introduction to the new branch of study called Logarithms.

It is a prerequisite here.

We also discussed about Logarithmic Function which is reproduced below.

Here we also discuss a few problems involving Logarithm Function.

Get The Best Grades With the Least Amount of Effort

Here is a collection of proven tips, tools and techniques to turn you into a super-achiever - even if you've never thought of yourself as a "gifted" student.

The secrets will help you absorb, digest and remember large chunks of information quickly and easily so you get the best grades with the least amount of effort.

If you apply what you read from the above collection, you can achieve best grades without giving up your fun, such as TV, surfing the net, playing video games or going out with friends!

Look at the following two sets of examples 1st set: 2^{3} = 8 ⇒ log_{2} 8 = 3 2^{4} = 16 ⇒ log_{2} 16 = 4 2^{5} = 32 ⇒ log_{2} 32 = 5 Here as the number (8, 16, 32) increases,the log of the number (3, 4, 5) increases. Note that the base (2) is more than 1. What is seen to be true here in this example, is in fact true in general. The general statement is

If a > 1, n_{1} > n_{2} ⇒ log_{a}n_{1} > log_{a}n_{2} i.e. when the base is more than 1, the logarithm function is an increasing function.

2nd set: (1⁄2)^{5} = 1⁄32 ⇒ log_{(1⁄2)} (1⁄32) = 5 (1⁄2)^{4} = 1⁄16 ⇒ log_{(1⁄2)} (1⁄16) = 4 (1⁄2)^{3} = 1⁄8 ⇒ log_{(1⁄2)} (1⁄8) = 3 Here as the number (1⁄32, 1⁄16, 1⁄8) increases,the logarithm of the number (5, 4, 3) decreases. Note that the base (1⁄2) is less than 1. What is seen to be true here in this example, is infact true in general. The general statement is

If 0 < a < 1, n_{1} > n_{2} ⇒ log_{a}n_{1} < log_{a}n_{2} i.e. when the base is less than 1 (and positive), the logarithm function is a decreasing function.

If log_{x}y = log_{y}z = log_{z}x, prove that x = y = z

Solution to Solved Example 1 of Logarithm Function :

Let log_{x}y = log_{y}z = log_{z}x = K By converting from Logarithmic Form to Exponential Form (See Formula 1), we get x^{K} = y........(i) y^{K} = z.............(ii) z^{K} = x.............(iii) Using the value of y from (i) in (ii), we get z = y^{K} = (x^{K})^{K} = x^{K x K} = x^{K2}.....(iv) Using the value of x from (iii) in (iv), we get z = x^{K2} = (z^{K})^{K2} = z^{K x K2} = z^{K3} Thus we have z^{1} = z^{K3} Since the bases are equal, the exponents have to be equal. ∴ K^{3} = 1 ⇒ K = 1 Using the value of K in (i) and (ii), we get x^{1} = y and y^{1} = z. ⇒ x = y = z (Proved.)

Solved Example 2 of Logarithm Function :

If (log x)⁄(b - c) = (log y)⁄(c - a) = (log z)⁄(a - b), then prove that xyz = x^{a}y^{b}z^{c} = x^{(b + c)}y^{(c + a)}z^{(a + b)} = 1

Solution to Solved Example 2 of Logarithm Function :

Let (log x)⁄(b - c) = (log y)⁄(c - a) = (log z)⁄(a - b) = K ⇒ log x = K(b - c); log y = K(c - a); log z = K(a - b); Let the base of the logarithms be p. Then the above three equations become log_{p}x = K(b - c); log_{p}y = K(c - a); log_{p}z = K(a - b) By converting from logarithmic form to exponential form (See Formula 1), we get p^{K(b - c)} = x; p^{K(c - a)} = y; p^{K(a - b)} = z; ⇒ xyz = p^{K(b - c)} x p^{K(c - a)} x p^{K(a - b)} = p^{{K(b - c) + K(c - a) + K(a - b)}} = p^{{K(b - c + c - a + a - b)}} = p^{{K(0)}} = p^{0} = 1. (Proved.)

We have p^{K(b - c)} = x; p^{K(c - a)} = y; p^{K(a - b)} = z; Raising the powers a, b, c to x, y, z respectively, we get x^{a} = (p^{K(b - c)})^{a} = p^{Ka(b - c)}; Similarly, y^{b} = p^{Kb(c - a)} and z^{c} = p^{Kc(a - b)} Multiplying these three, we get x^{a}y^{b}z^{c} = p^{Ka(b - c)}p^{Kb(c - a)}p^{Kc(a - b)} = p^{{Ka(b - c) + Kb(c - a) + Kc(a - b)}}= p^{K{ab - ac + bc - ba + ca - cb}} = p^{K{0}} = p^{0} = 1. (Proved.)

Raising the powers (b + c), (c + a), (a + b) to x, y, z respectively, we get x^{(b + c)} = {p^{K(b - c)}}^{(b + c)} = p^{K(b + c)(b - c)} = p^{K(b2 - c2)}; Similarly, y^{(c + a)} = p^{K(c2 - a2)} and z^{(a + b)} = p^{K(a2 - b2)} Multiplying these three, we get x^{(b + c)}y^{(c + a)}z^{(a + b)} = p^{K(b2 - c2)}p^{K(c2 - a2)}p^{K(a2 - b2)} = p^{K(b2 - c2) + K(c2 - a2) + K(a2 - b2)} = p^{K(b2 - c2 + c2 - a2 + a2 - b2)} = p^{K(0)} = p^{0} = 1.(Proved.)

Thus, the problem on Logarithm Function is solved.

More Solved Examples : Logarithm Function

The following Link takes you to more Solved Examples.

Are you spending lot of money for math tutors to your child and still not satisfied with his/her grades ?

Do you feel that more time from the tutor and more personalized Math Help to identify and fix the problems faced by your child will help ?

Here is a fool proof solution I strongly recommend and that too With a minuscule fraction of the amount you spent on tutors with unconditional 100% money back Guarantee, if you are not satisfied.

It is like having an unlimited time from an excellent Tutor.

It is an Internet-based math tutoring software program that identifies exactly where your child needs help and then creates a personal instruction plan tailored to your child’s specific needs.

If your child can use a computer and access the Internet, he or she can use the program. And your child can access the program anytime from any computer with Internet access.

There is an exclusive, Parent Information Page provides YOU with detailed reports of your child’s progress so you can monitor your child’s success and give them encouragement. These Reports include

Time spent using the program

Assessment results

Personalized remediation curriculum designed for your child

Details the areas of weakness where your child needs additional help

Provides the REASONS WHY your child missed a concept

List of modules accessed and amount of time spent in each module

Quiz results

Creates reports that can be printed and used to discuss issues with your child’s teachers

These reports are created and stored in a secure section of the program, available exclusively to you, the parent. The section is accessed by a password that YOU create and use. No unauthorized users can access this information.

Its research-based results have proven that it really works for all students! in improving math skills and a TWO LETTER GRADE INCREASE in math test scores!,if they invest time in using the program.

Proven for More than 10,000 U.S. public school students who increased their math scores.

Solve the following problems on Logarithm Function.

If a^{x} = b^{y} = c^{z} and y^{2} = zx, Prove that log_{b}a = log_{c}b.

Find the value of x from the equation a^{(3 - x)}.b^{5x} = a^{3x}.b^{(x + 5)}

For Answers to these problems on Logarithm Function see at the bottom of the page.

Answers to Exercise : Logarithm Function

(2) (5 log b - 3 log a)⁄{4(log b - log a)}

Progressive Learning of Math : Logarithm Function

Recently, I have found a series of math curricula (Both Hard Copy and Digital Copy) developed by a Lady Teacher who taught everyone from Pre-K students to doctoral students and who is a Ph.D. in Mathematics Education.

This series is very different and advantageous over many of the traditional books available. These give students tools that other books do not. Other books just give practice. These teach students “tricks” and new ways to think.

These build a student’s new knowledge of concepts from their existing knowledge. These provide many pages of practice that gradually increases in difficulty and provide constant review.

These also provide teachers and parents with lessons on how to work with the child on the concepts.

The series is low to reasonably priced and include