Write 2/3 log 27 − 3 log 2 + 3/2 log 25 as a single logarithm
Answers
Answer:
1. Find the logarithms of:
(i) 1728 to the base 2√3
Solution:
Let x denote the required logarithm.
Therefore, log2√3 1728 = x
or, (2√3)x = 1728 = 26 ∙ 33 = 26 ∙ (√3)6
or, (2√3)x = (2√3)6
Therefore, x = 6.
(ii) 0.000001 to the base 0.01.
Solution:
Let y be the required logarithm.
Therefore, log0.01 0.000001 = y
or, (0.01y = 0.000001 = (0.01)3
Therefore, y = 3.
2. Proof that, log2 log2 log2 16 = 1.
Solution:
L. H. S. = log2 log2 log2 24
= log2 log2 4 log2 2
= log2 log2 22 [since log2 2 = 1]
= log2 2 log2 2
= 1 ∙ 1
= 1. Proved.
3. If logarithm of 5832 be 6, find the base.
Solution:
Let x be the required base.
Therefore, logx 5832 = 6
or, x6 = 5832 = 36 ∙ 23 = 36 ∙ (√2)6 = (3 √2)6
Therefore, x = 3√2
Therefore, the required base is 3√2
4. If 3 + log10 x = 2 log10 y, find x in terms of y.
Solution:
3 + log10 x = 2 log10 y
or, 3 log10 10 + log10 x= 1og10 y2 [since log10 10 = 1]
or. log10 103 + log10 x = log10 y2
or, log10 (103 ∙ x) = log10 y2
or, 103 x = y2
or, x = y2/1000, which gives x in terms y.
5. Prove that, 7 log (10/9) + 3 log (81/80) = 2log (25/24) + log 2.
Solution:
Since,7 log (10/9) + 3 log (81/80) - 2 log (25/24)
= 7(log 10 – log 9)+ 3(1og 81 - log 80)- 2(1og 25 - 1og 24)
= 7[log(2 ∙ 5) - log32] + 3[1og34 - log(5 ∙ 24)] - 2[log52 - log(3 ∙ 23)]
= 7[log 2 + log 5 – 2 log 3] + 3[4 log 3 - log 5 - 4 log 2] - 2[2 log 5 – log 3 – 3 log 2]
= 7 log 2+ 7 log 5 - 14 log 3 + 12 log 3 – 3 log 5 – 12 log 2 – 4 log 5 + 2 log 3 + 6 log 2
= 13 log 2 – 12 log 2 + 7 log 5 – 7 log 5 – 14 log 3 + 14 log 3 = log 2
Therefore 7 log(10/9) +3 log (81/80) = 2 log (25/24) + log 2. Proved.
6. If log10 2 = 0.30103, log10 3 = 0.47712 and log10 7 = 0.84510, find the values of
(i) log10 45
(ii) log10 105.
(i) log10 45
Solution:
log10 45 = log10 (5 × 9)
= log10 5 + log10 9
= log10 (10/2) + log10 32
= log10 10 - log10 2 + 2 log10 3
= 1 - 0.30103 + 2 × 0.47712
= 1.65321.
(ii) log10 105
Solution:
log10 105
= log10 (7 x 5 x 3)
= log10 7 + log10 5 + log10 3
= log10 7 + log10 10/2 + log10 3
= log10 7 + log10 10 - log10 2 + log10 3
= 0.845l0 + 1 - 0.30103 + 0.47712
= 2.02119.
7. Prove that, logb a × logc b × logd c = logd a.
Solution:
L. H. S. = logb a × logc b × logd C
= logc a × logd c [since logb M × loga b = loga M]
= logd a. (using the same formula)
Alternative Method:
Let, logb a = x Since, bx = a,
logc b = y Therefore, cy = b
and logd c = z Therefore, dz = c.
Now, a = bx = (cy)x = cxy = (dz)xy = dxyz
Therefore logd a = xyz = logb a × logc b × logd c. (putting the value of x, y, z)
8. Show that, log4 2 × log2 3= log4 5 × log5 3.
Solution:
L. H. S. = log4 2 × log2 3
= log4 3
= log5 3 × log4 5. Proved.
9. Show that, log2 10 - log8 125 = 1.
Solution:
We have, log8 125 = log8 53 = 3 log8 5
= 3 ∙ (1/log5 8) = 3 ∙ (1/log5 23) = 3 ∙ (1/3 log5 2) = log2 5
Therefore, L.H. S. = log2 10 - log8 125 = log2 10 - log2 5
= log2 (10/5) = log2 2 = 1. Proved.
10. If log x/(y - z) = log y/(z - x) = log z/(x – y)
show that, xx yy zz = 1
Solution:
Let, log x/(y - z) = log y/(z - x) = log z/(x – y) = k
Therefore, log x = k(y - z) ⇔ x log x = kx(y - z )
or, log xx = kx(y - z) ... (1)
Similarly, log yy = ky (z - x) ... (2)
and log zz = kz(x - y) ... (3)
Now, adding (1), (2) and (3) we get,
log xx + log yy + log zz = k (xy - xz + yz - xy + zx - yz)
or, log (xx yy zz) = k × 0 = 0 = log 1
Therefore, xx yy zz = 1 Proved.
11. If a2 - x ∙ b5x = ax + 3 ∙ b3x show that, x log (b/a) = (1/2) log a.
Solution:
a2 - x ∙ b5x = ax + 3 ∙ b3x
Therefore, b5x/b 3x = ax + 3/a 2 - x
or, b5x - 3x = ax + 3 – 2 + x
or, b 2x = a2x + 1 or, b 2x =a 2x ∙ a
or, (b/a)2x = a
or, log (b/a)2x = log a (taking logarithm both sides)
or, 2x log (b/a) =log a
or, x log (b/a) = (1/2) log a Proved.
12. Show that, aloga2 x × blog b2 y × clog c2 z = √xyz
Solution:
Let, p = alog a2 x
Now, taking logarithm to the base a of both sides we get,
loga p = loga alog a2 x
⇒ loga p = loga2 x ∙ loga a
⇒ loga p = loga2 x [since, loga a = 1]
⇒ loga p = 1/(logx a2) [since, logn m = 1/(logm n)]
⇒ loga p = 1/(2 logx a)
⇒ loga p = (1/2) loga x
⇒ loga p = loga x ½
⇒ loga p = loga √x
Therefore, p = √x or, aloga2 x = √x
Similarly, blogb2 y = √y and clogc2 z = √z
L.H.S = √x ∙ √y ∙ √z = √xyz Proved.
13. If y = a1/(1 – loga x) and z = a1/(1 – loga y) show that, x = a1/(1 – loga z)
Solution:
Let, loga x = p, loga y = q and loga z = r
Then, by problem, y = a1/(1 - p) ...……….. (1)
and z = a1/(1 - q) .............. (2)
Now, taking logarithm to the base a of both sides of (1) we get,
loga y = loga a1/(1 - p)
or, q = 1/(1 – p), [since loga a = 1]
Again, taking logarithm to the base a of both sides of (2) we get,
loga z = loga a1/(1 - q)
or, r = 1/(1 – q)
or, 1 - q = 1/r
or, 1 - 1/(1 – p) = 1/r
or, 1 - 1/r = 1/(1 – p)
or, (r – 1)/r = 1/(1 – p)
or, 1 - p = r/(r – 1)
or, p = 1- r/(r – 1) = 1/(1 – r)
or, loga x = 1/(1-loga z)
or, x = a1/(1 – loga z) Proved.
14. If x, y,z are in G. P., prove that, loga x+ loga z = 2/(logy a )[x, y, z, a > 0).
Step-by-step explanation:
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