Logarithm Solved Questions and Answers (original) (raw)
Last Updated : 11 Jul, 2025
A logarithm is the exponent to which a base must be raised to obtain a given number. The logarithmic function is the inverse of the exponential function. It is defined as:
y = logax, if and only if x = ay; for x > 0, a > 0, and a ≠ 1.
**Logarithm questions and answers are provided below for you to learn and practice.
**Question 1: Find the value of x in the equation given 8x+1 – 8x-1 = 63.
**Solution:
we know that: 8x+1 = 8⋅8x and 8x-1 = 8x/ 8
So the equation becomes: 8⋅8x − 8x/8 = 63
Let y=8x
Substitute: 8⋅y − y/8 = 63
(64y - y)/8 = 63
63y /8 = 63
y = 504/63 = 8
8x = 81
**x = 1
**Question 2: Find the value of x for the equation given log0.25x = 16.
**Solution:
log0.25x = 16
It can be write as
x = (0.25)16
x = (1/4)16
**x = 4 -16
**Question 3: Solve the equation log121728 x log96561.
**Solution:
It can be written as:
log12(123) x log9(94)
= 3log1212 x 4log99
= 3 x 4 = **12
**Question 4: Solve for x : logx3 + logx9 + logx27 + logx81 = 10.
**Solution:
It can be write as:
logx(3 x 9 x 27 x 81) = 10
logx(31 x 32 x 33 x 34) = 10
logx(310) = 10
10 logx3 = 10
then, **x = 3
**Question 5: If log(a + 3 ) + log(a – 3) = 1 ,then a=?
**Solution:
log10((a + 3)(a – 3))=1
log10(a2 – 9) = 1
(a2 – 9) = 10
a2 = 19
**a = √19
**Question 6: Solve 1/logab(abcd) + 1/logbc(abcd) + 1/logcd(abcd) + 1/logda(abcd).
**Solution:
=logabcd(ab) + logabcd(bc) + logabcd(cd) + logabcd(da)
=logabcd(ab × bc × cd × da)
=logabcd(abcd)2
=2 logabcd(abcd)
=**2
**Question 7: If xyz = 10, then solve log(xn yn / zn) + log(yn zn / xn) + log(zn xn / yn).
**Solution:
log(xn yn / zn * yn zn / xn * zn xn / yn)
= log xn yn zn
= log(xyz)n
= n log(xyz)
= n log(10)
= n×1
**= n
**Question 8: Find (121/10) x = 3.
**Solution:
Apply logarithm on both sides
log(121/10)(121/10)x = log(121/10)3
x = (log 3) / (log 121 – log 10)
**x = (log 3) / (2 log 11 – 1)
**Question 9: Solve log(2x2 + 17)= log (x – 3)2.
**Solution:
log(2x2 + 17)= log (x2 – 6x + 9)
2x2 + 17 = x2 – 6x + 9
x2 + 6x + 8 = 0
x2 + 4x + 2x + 8 = 0
x(x + 4) + 2(x + 4) = 0
(x + 4)(x + 2)=0
**x= -4,-2
**Question 10: log2(33 – 3x)= 10log(5 – x). Solve for x.
**Solution:
Put x = 0
log2(33 – 1)= 10log(5)
log232 = 5
5 log2 2 = 5
5 = 5
LHS = RHS