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

Please study

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 ⇒ log_a n₁ > log_a 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 ⇒ log_a n₁ < log_a n₂ i.e. when the base is less than 1 (and positive), the logarithm function is a decreasing function.

Solved Example 1 of Logarithm 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 z¹ = z^K3
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 = x^ay^bz^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)} = p⁰ = 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^ay^bz^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} = p⁰ = 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) = p⁰ = 1.(Proved.)

Thus, the problem on Logarithm Function is solved.

More Solved Examples : Logarithm Function

The following Link takes you
to more Solved Examples.

More Solved Examples

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Exercise : Logarithm Function

Solve the following problems on Logarithm Function.

1. If a^x = b^y = c^z and y² = zx,
   Prove that log_b a = log_c b.
   
2. 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)}

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