# WORKSHEET ON PROPERTIES OF LOGARITHMS

Question 1 :

If a2 + b2 = 7ab, show that :

log(a+b)/3  =  1/2(log a+log b)

Question 2 :

Prove :

log(a2/bc) + log (b2/ac) + log (c2/ab)  =  0

Question 3 :

Prove :

log 2 + 16 log (16/15) + 12 log (25/24) + 7 log(81/80)  =  1

Question 4 :

Prove :

Question 5 :

Prove :

log a + log a2 + log a3 + · · · + log an = (n(n+1)/2) log a

Question 6 :

If log x/(y - z)  =  log y/(z - x)  =  log z/(x - y), then prove :

xyz  =  1

Question 7 :

Solve for x :

log2x − 3log1/2x  =  6

Question 8 :

Solve for x :

log5-x(x2 − 6x + 65)  =  2

Question 1 :

If a2 + b2 = 7ab, show that :

log(a+b)/3  =  1/2(log a+log b)

Given that :

a2 + b2 = 7ab

a2 + b2 + 2ab = 7ab + 2ab

(a + b)2  =  9ab

(a + b)  =  √(9ab)  ==>  a + b  =  3(ab)

(a + b)/3  =  (ab)1/2

Take log on both sides

log (a + b)/3  =  log (ab)1/2

log (a + b)/3  =  (1/2)(log a + log b)

Hence proved.

Question 2 :

Prove :

log(a2/bc) + log (b2/ac) + log (c2/ab)  =  0

L.H.S  =  log (a2/bc) + log (b2/ac) + log (c2/ab)

L.H.S  =  log (a2/bc) ⋅ (b2/ac)  (c2/ab)

L.H.S  =  log 1

L.H.S  =  0

L.H.S  =  R.H.S

Hence proved.

Question 3 :

Prove :

log 2 + 16 log (16/15) + 12 log (25/24) + 7 log(81/80)  =  1

L.H.S

log 2 + 16 log (16/15) + 12 log (25/24) + 7 log(81/80)  =  1

=  log 10

=   1  =  R.H.S

Hence proved

Question 4 :

Prove :

L.H.S

Let us use base change rule,

=  (1/loga a2 (1/logb b2 (1/logc c2)

=  (1/2logaa)  (1/2logbb)  (1/2logcc)

=  (1/2) ⋅ (1/2) ⋅ (1/2)

=  1/8

Hence proved

Question 5 :

Prove :

log a + log a2 + log a3 + · · · + log an(n(n+1)/2) log a

L.H.S  =  log a + log a2 + log a3 + · · · + log an

L.H.S  =  1log a + 2log a + 3log a + · · · + nlog a

L.H.S  =  log a (1 + 2 + 3 + · · · + n)

L.H.S  =  log a (n(n + 1)/2)

L.H.S  =  [n(n+1)/2] ⋅ log a

L.H.S  =  R.H.S

Hence proved.

Question 6 :

If log x/(y - z)  =  log y/(z - x)  =  log z/(x - y), then prove :

xyz  =  1

 log x/(y-z)  =  k log y/(z-x)  =  k log z/(x-y)  =  k

log x  =  k(y-z)  == > ky - kz   ------(1)

log y  =  k(z-x) == > kz - kx   ------(2)

log z  =  k(x-y) == > kx - ky   ------(3)

(1) + (2) + (3)

log x + log y + log z =  ky - kz + kz + kx + kx - ky

log x + log y + log z =  0

log (xyz)  =  0

xyz  =  1

Hence proved.

Question 7 :

Solve for x :

log2x − 3log1/2x  =  6

logx − 3log 1/2 x  =  6

[1 / logx 2]  −  [3 / logx 1/2]  =  6

[1 / logx 2]  −  [3 / (logx 1 - logx 2)]  =  6

[1 / logx 2]  −  [3 / (0 - logx 2)]  =  6

[1 / logx 2]  −  [3 / (- logx 2)]  =  6

[1 / logx 2]  −  [- 3 / logx 2]  =  6

[1 / logx 2]  +  [3 / logx 2]  =  6

(1 + 3) / logx 2  =  6

4 / logx 2  =  6

4log2x  =  6

Divide both sides by 4.

log2x  =  6 / 4

log2x  =  3 / 2

x  =  23/2

x  =  2√2

Hence, the value of x is 2√2.

Question 8 :

Solve for x :

log5-x(x2 − 6x + 65)  =  2

log5-x (x2 − 6x + 65)  =  2

x2 − 6x + 65  =  (5 - x)2

x2 − 6x + 65  =  52 - 2 ⋅ ⋅ 5 + x2

x2 − 6x + 65  =  25 - 10x + x2

Simplify.

4x  =  - 40

Divide both sides by 4.

x  =  - 10

Hence, the value of x is -10.

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