# LOGARITHM FUNCTION - NATURE OF THE FUNCTION, SOLVED EXAMPLES AND EXERCISES

Introduction to Logarithms before Logarithm Function

if you have not already done so.

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.

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### Nature of Logarithmic Function :

Look at the following two sets of examples
1st set:
23 = 8 ⇒ log2 8 = 3
24 = 16 ⇒ log2 16 = 4
25 = 32 ⇒ log2 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, n1 > n2 ⇒ loga n1 > loga n2
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, n1 > n2 ⇒ loga n1 < loga n2 i.e. when the base is less than 1 (and positive), the logarithm function is a decreasing function.

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### Solved Example 1 of Logarithm Function :

If logx y = logy z = logz x, prove that x = y = z

Solution to Solved Example 1 of Logarithm Function :

Let logx y = logy z = logz x = K
By converting from Logarithmic Form to Exponential Form (See Formula 1), we get
xK = y........(i)
yK = z.............(ii)
zK = x.............(iii)
Using the value of y from (i) in (ii), we get
z = yK = (xK)K = xK x K = xK2.....(iv)
Using the value of x from (iii) in (iv), we get
z = xK2 = (zK)K2
= zK x K2 = zK3
Thus we have z1 = zK3
Since the bases are equal, the exponents have to be equal.
∴ K3 = 1 ⇒ K = 1
Using the value of K in (i) and (ii), we get x1 = y and y1 = 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 = xaybzc = 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
logp x = K(b - c); logp y = K(c - a); logp z = K(a - b)
By converting from logarithmic form to exponential form (See Formula 1), we get
pK(b - c) = x; pK(c - a) = y; pK(a - b) = z;
xyz = pK(b - c) x pK(c - a) x pK(a - b)
= p{K(b - c) + K(c - a) + K(a - b)} = p{K(b - c + c - a + a - b)}
= p{K(0)} = p0 = 1. (Proved.)

We have pK(b - c) = x; pK(c - a) = y; pK(a - b) = z;
Raising the powers a, b, c to x, y, z respectively, we get
xa = (pK(b - c))a
= pKa(b - c);
Similarly, yb = pKb(c - a)
and zc = pKc(a - b)
Multiplying these three, we get
xaybzc = pKa(b - c)pKb(c - a)pKc(a - b) = p{Ka(b - c) + Kb(c - a) + Kc(a - b)}= pK{ab - ac + bc - ba + ca - cb}
= pK{0} = p0 = 1. (Proved.)

Raising the powers (b + c), (c + a), (a + b) to x, y, z respectively, we get
x(b + c) = {pK(b - c)}(b + c)
= pK(b + c)(b - c) = pK(b2 - c2);
Similarly, y(c + a) = pK(c2 - a2)
and z(a + b) = pK(a2 - b2)
Multiplying these three, we get
x(b + c)y(c + a)z(a + b)
= pK(b2 - c2)pK(c2 - a2)pK(a2 - b2)
= pK(b2 - c2) + K(c2 - a2) + K(a2 - b2)
= pK(b2 - c2 + c2 - a2 + a2 - b2)
= pK(0) = p0 = 1.(Proved.)

Thus, the problem on Logarithm Function is solved.

### More Solved Examples : Logarithm Function

to more Solved Examples.

More Solved Examples

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### Exercise : Logarithm Function

Solve the following problems on Logarithm Function.

1. If ax = by = cz and y2 = zx,
Prove that logb a = logc b.
2. Find the value of x from the equation
a(3 - x).b5x = a3x.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)}

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