CHAPTER 3: LOGARITHM
3.1.4. Laws of Logarithm
(1) logamn = logam + logan
Proof:
Let, logam = x and logan = y
Now,
EXERCISE 3.1
1. Find the values
(i) log66
Solution:
log66
= 1
(ii) log0.40.4
Solution:
log0.40.4
= 1
(iii) log100.1
Solution:
log100.1
= log1010–1
= (– 1)log1010
= – 1
(iv) log10010
Solution:
log10010
(v) log4256
Solution:
log4256
= log444
= 4log44
= 4
(vi)
Solution:
(vii) log25125
Solution:
log25125
= log2553
= 3log255
(viii)
Solution:
(ix) log0.110–4
Solution:
log0.110–4
= – 4log0.110
(x)
Solution:
2. Find the value of x, if
(i)
Solution:
⇒ 23 = x
⇒ x = 8
(ii) logx16 =2
Solution:
logx16 = 2
⇒ x2 = 16
⇒ x = ±4
(iii) log3x = – 3
Solution:
log3x = – 3
⇒ 3–3 = x
(iv)
Solution:
(v)
Solution:
(vi) log100.01 = x
Solution:
log100.01 = x
3. (i) Find the sum of and .
Solution:
(ii) If log23 = p, then find the values of log827 and log1681.
Solution:
Given,
Now, log827 = log833
and
= p
(iii) If , then find the value of ,and
Solution:
Given,
Now,
4. Verify:
(i)
Solution:
⸫LHS = RHS
(ii)
Solution:
5. (i) Show that
Solution:
(ii) If , show that x = 8.
Solution:
Given,
6. (i) Prove that
Solution:
(ii)
Solution:
(iii)
Solution:
(iv)
Solution:
(v)
Solution:
(vi)
Solution:
(vii)
Solution:
(viii)
Solution:
7. Simplify:
(i) log35 × log2527
Solution:
log35 × log2527
(ii) log210 – log8125
Solution:
log210 – log8125
(iii)
Solution:
(iv)
Solution:
(v)
Solution:
(vi)
Solution:
8. Show that:
(i)
Solution:
(ii)
Solution:
(iii)
Solution:
(iv)
Solution:
(v)
Solution:
(vi)
Solution:
(vii)
Solution:
Now,
9. (i) If then prove that
Solution:
Given,
…………. (1)
Similarly,
…………. (2)
…………. (3)
(1)×(2)×(3)⇒
(ii) If then prove that
Solution:
Given,
(iii) If then prove that
Solution:
Given,
Similarly,
and
Now,
(iv) If , then what will be the value of ? Also, show that
Solution:
Given,
Now,
(v) If , then find the value of abc. Also prove that aabbcc = 1, a, b, c > 0
Solution:
Let,
Now,
Again,
(vi) If a4 + b4 = 14a2b2, then show that log(a2 + b2) = loga + logb + 2log2, (a, b > 0).
Solution:
Given,
a4 + b4 = 14a2b2
⇒ a4 + b4 + 2a2b2= 14a2b2 + 2a2b2
⇒ (a2 + b2)2 = 16a2b2
⇒ log(a2 + b2)2 = log(16a2b2)
⇒ 2log(a2 + b2) = log24 + loga2 + logb2
⇒ 2log(a2 + b2) = 4log2 + 2loga + 2logb
⇒ log(a2 + b2) = loga + logb + 2log2
(vii), then what will be the value of xyz?
Solution:
Given,
Now,
10. If , then prove that,
(i)
Solution:
Let,
Now,
(ii)
Solution:
Let,
Now,
(iii)
Solution:
11. (i) If logxy = 6 and log14x8y = 3, then x = ?
Solution:
Given,
logxy = 6
⇒ x6 = y ………………. (1)
and
log14x8y = 3
⇒ (14x)3 = 8y
⇒ 2744x3 = 8x6 [From (1)]
⇒ 343 = x3
⇒ 73 = x3
⇒ x = 7
(ii) If logab = 6 and log6a27b = 3, then find the value of a.
Solution:
Given,
logab = 6
⇒ a6 = b …………….. (1)
and
log6a27b = 3
⇒ (6a)3 = 27b
⇒ 216a3 = 27b
⇒ 216a3 = 27a6 [From (1)]
⇒ 8 = a3
⇒ 23 = a3
⇒ a = 2
(iii) 5log10a + log10b = 2 then show that b = 100a–5.
Solution:
Given,
5log10a + log10b = 2
⇒ log10a5 + log10b = 2log1010
⇒ log10(a5b) = log10102
⇒ a5b = 100
⇒ b = 100a–5
12. Solve:
(i) log2x + log2(x + 6) = 4
Solution:
log2x + log2(x + 6) = 4
⇒ log2x(x + 6) = 4log22
⇒ log2x(x + 6) = log224
⇒ x(x + 6) = 16
⇒ x2 + 6x – 16 = 0
⇒ x2 + 8x – 2x – 16 = 0
⇒ x(x + 8) – 2(x + 8) = 0
⇒ (x + 8)(x – 2) = 0
⇒ x + 8 = 0 OR x – 2= 0
⇒ x = – 8 OR x = 2
(ii) log(1 + x) = 2logx
Solution:
log(1 + x) = 2logx
⇒ log(1 + x) = logx2
⇒ 1 + x = x2
⇒ x2 – x – 1 = 0
(iii) logx(8x – 3) = 2 + logx4
Solution:
logx(8x – 3) = 2 + logx4
⇒ logx(8x – 3) = 2logxx + logx4
⇒ logx(8x – 3) = logxx2 + logx4
⇒ logx(8x – 3) = logx4x2
⇒ 8x – 3 = 4x2
⇒ 4x2 – 8x + 3 = 0
⇒ 4x2 – 6x – 2x + 3 = 0
⇒ 2x(2x – 3) –1(2x – 3) = 0
⇒ (2x – 3) (2x – 1) = 0
⇒ 2x – 3 = 0 OR 2x – 1 = 0
⇒ 2x = 3 OR 2x = 1
(iv) log3x + log9x + log27x =
Solution:
log3x + log9x + log27x =
(v) logx3 + logx9 + logx729 = 9
Solution:
logx3 + logx9 + logx729 = 9
⇒ logx(3 × 9 ×729) = 9
⇒ logx(3×32×36) = 9
⇒ logx39 = 9
⇒ 9logx3 = 9
⇒ logx3 = 1
⇒ x1 = 3
⇒ x = 3
(vi) 5logx – 3logx – 1 = 3logx + 1 – 5logx – 1
Solution:
5logx – 3logx – 1 = 3logx + 1 – 5logx – 1
⇒ 5logx + 5logx – 1 = 3logx + 1 + 3logx – 1
⇒ 5logx + 5logx×5– 1 = 3logx×3 + 3logx×3– 1
⇒ 5logx(1 + 5– 1) = 3logx(3 + 3– 1)
⇒ logx = 2
⇒ 102 = x
⇒ x = 100
(vii)
Solution:
(viii) logx(3x2 + 10x) = 3
Solution:
logx(3x2 + 10x) = 3
⇒ x3 = 3x2 + 10x
⇒ x3 – 3x2 – 10x = 0
⇒ x(x2 – 3x – 10) = 0
⇒ x(x2 – 5x + 2x – 10) = 0
⇒ x{x(x – 5) + 2(x – 5)} = 0
⇒ x(x – 5)(x + 2) = 0
⇒ x = 0 OR x – 5 = 0 OR x + 2 = 0
⇒ x = 0 OR x = 5 OR x = – 2
⸪x < 0 is not possible.
⸫x = 5
(ix) log(3x + 2) +log(3x – 2) = 5log2
Solution:
log(3x + 2) +log(3x – 2) = 5log2
⇒ log(3x + 2)(3x – 2) = log25
⇒ (3x + 2)(3x – 2) = 32
⇒ 9x2 – 4 = 32
⇒ 9x2 = 36
⇒ x2 = 4
⇒ x = ±2
13. If al = bm = cn = dp, then show that
Solution:
Let,
al = bm = cn = dp = k
Now,
14. Prove that:
(i)
Solution:
We know that,
(ii)
Solution:
We know that,
(iii)
Solution:
We know that,
