Worksheet on Properties of Logarithms :
Worksheet given in this section will be much useful for the students who would like to practice solving problems on properties of logarithms.
Before look at the worksheet, if you would like to learn the 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)
Solution :
Given that :
a2 + b2 = 7ab
Add 2ab on both sides
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
Solution :
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
Solution :
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 :
Solution :
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
Solution :
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
Solution :
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
Solution :
log2 x − 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
Solution :
log5-x (x2 − 6x + 65) = 2
x2 − 6x + 65 = (5 - x)2
x2 − 6x + 65 = 52 - 2 ⋅ x ⋅ 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.
After having gone through the stuff given above, we hope that the students would have understood how to solve problems using properties of logarithms.
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