Question

Differentiate $\tan^{-1}( \frac{a-x}{1+ax})$ w.r.t $\sin^{-1}( 3x-4x^{3})$

Answer 1

(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 log107 = 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). 

Solution: 

By problem, x, y, z are in G. P. 

Therefore, y/x = z/y or, zx = y2