Logarithm and Exponential Questions with Answers and Solutions
Questions on Logarithm and exponential with answers and detailed solutions, for grade 11, are presented.
Review
1) One of the most important property of logarithmic and exponential functions is that they are inverse of each other and therefore we can convert exponential and logarithmic expressions using the following:
Numerical example:
a) If b^{ x} = b^{ y} then x = y
b) If log_{b} x = log_{b} y then x = y
Answer the Questions

Change the given logarithmic expressions into exponential expressions:
a) log_{x} (a) = c
b) log_{b} (2x + 1) = 3

Change the given exponential expressions into logarithmic expressions:
a) 3^{x} = m
b) x^{2} = a

Evaluate , without calculator, he following logarithmic expressions:
a) log_{2} 16
b) log_{3} 27
c) log_{2} (1/32)
d) log_{25} 5
e) Log √(10)
f) log_{b} 1 , with b > 0 and b ≠ 0
g) log_{0.1} 10

Solve for x the following logarithmic equations:
a) log_{2} x = 3
b) log_{x} 8 = 3
c) log_{3} x = 1
d) log_{5.6} x = 0
e) log_{2} (3x + 1) = 4
f) log_{3} (1/(x + 1)) = 2
g) log_{4} ( (x + 1)/(2x  1) ) = 0
h) Log ( 1/x + 1 ) = 2
i) log_{x} 0.0001 = 4

Solve for x the following exponential equations:
a) 3 ^{x} = 9
b) 4 ^{2x + 1} = 16
c) (1 / 2) ^{x} = 2
d) 10 ^{x} = 5
e) (1 / 3) ^{x/2  2} = 9
f) 0.01 ^{x} = 100
g) 2^{2x}  6(2^{x}) =  8
Solutions to the Above Questions

Solution
Use the equivalent expressions : y = log _{b} (x) ⇔ x = b^{ y} to write
a) log_{x} (a) = c as an exponential a = x^{ c}
b) log_{b} (2x + 1) = 3 as an exponential 2x + 1 = b^{ 3}

Solution
Use the equivalent expressions : x = b^{ y} ⇔ y = log _{b} (x) to write
a) 3^{x} = m as a logarithm x = log _{3} (m)
b) x^{2} = a as a logarithm 2 = log _{x} (a)

Solution
Use the equivalent expressions : y = log _{b} (x) ⇔ x = b^{ y} evaluate the following without calculator:
a) let y = log_{2} 16
convert to exponential form: 2^{ y} = 16 = 2^{ 4}, which gives 2^{ y} = 2^{ 4}
hence using property 2 a) above " If 2^{ y} = 2^{ 4} then y = 4"
hence : y = log_{2} 16 = 4
b) let y = log_{3} 27
convert to exponential form: 3^{ y} = 27 = 3^{ 3}, which gives 3^{ y} = 3^{ 3}
hence using property 2 a) above " If 3^{ y} = 3^{ 3} then y = 3"
hence : y = log_{3} 27 = 3
c) let y = log_{2} (1/32)
convert to exponential form: 2^{ y} = 1 / 32 = 1 /(2^{ 5}) = 2^{ 5}, which gives 2^{ y} = 2^{ 5}
2^{ y} = 2^{ 5} gives y = 5"
hence y = log_{2} (1/32) = 5
d) let y = log_{25} 5
convert to exponential form: 25^{ y} = 5 = √(25) = 25^{ 1/2} which gives 25^{ y} = 25^{ 1/2}
hence y = log_{25} 5 = 1/2
e) Let y = Log √(10) ; convert to exponential form: 10^{ y} = √(10) = 10^{ 1/2}, hence y = Log √(10) = 1/2
f) Let y = log_{b} 1 , ( with b > 0 and b ≠ 0) ; convert to exponential form: b^{ y} = 1 = b^{ 0}, hence y = log_{b} 1 = 0
g) let y = log_{0.1} 10 ; convert to exponential form: 0.1^{ y} = 10 = 1 / 0.1 = (0.1)^{ 1} , hence y = log_{0.1} 10 = 1

Solution
Use the equivalent expressions : y = log _{b} (x) ⇔ x = b^{ y} to solve for x the following logarithmic equations:
a) log_{2} x = 3 ; convert to exponential form: x = 2 ^{ 3} = 8
b) log_{x} 8 = 3 ; convert to exponential form: 8 = x ^{ 3} , write 8 as 8 = 2 ^{ 3} ; hence x = 2
c) log_{3} x = 1 ; convert to exponential form: x = 3 ^{ 1} = 3
d) log_{5.6} x = 0 ; convert to exponential form: x = 5.6 ^{ 0} = 1
e) log_{2} (3x + 1) = 4 ; convert to exponential form: 3x + 1 = 2 ^{ 4} = 16 , solve for x: 3x + 1 = 16 , 3x = 15 , x = 5
f) log_{3} (1/(x + 1)) = 2 ; convert to exponential form: 1/(x + 1) = 3 ^{ 2} = 9 , solve for x: 1/(x + 1) = 9 , 1 = 9x + 9 , x =  8 / 9
g) log_{4} ( (x + 1)/(2x  1) ) = 0 ; convert to exponential form: (x + 1)/(2x  1) = 4 ^{ 0} = 1 , solve for x: (x + 1)/(2x  1) = 1 , x + 1 = 2x  1 , x = 2
h) Log ( 1/x + 1 ) = 2 ; convert to exponential form: 1/x + 1 = 10 ^{ 2} = 100 , solve for x: 1/x + 1 = 100 , 1/x = 99 , x = 1/99
i) log_{x} 0.0001 = 4 ; convert to exponential form: x ^{ 4} = 0.0001 = 1/10000 = 1/10^{4} = (1/10)^{4}
which gives x ^{ 4} = (1/10)^{4}
hence x = 1/10.

Solution
Use the the property: if b^{ x} = b^{ y} then x = y to solve the exponential functions:
Note that in the above equation, the bases of the two exponential are both equal to a.
a) 3 ^{x} = 9 = 3 ^{2} , hence x = 2
b) 4 ^{2x + 1} = 16 = 4^{2}, which gives 4 ^{2x + 1} = 4^{2}, hence 2x + 1 = 2 , x = 1 / 2
c) (1 / 2) ^{x} = 2 = 1 / 2^{ 1} = (1/2)^{ 1}, which gives (1 / 2) ^{x} = (1/2)^{ 1} , hence x =  1
d) 10 ^{x} = 5 , convert to logarithm: x = log _{10} 5 = Log 5 (here we have used convertion to solve the given equation)
e) (1 / 3) ^{x/2  2} = 9 = 3 ^{ 2} = (1 / 3^{ 2}) = (1 / 3)^{ 2} , which gives (1 / 3) ^{x/2  2} = (1 / 3)^{ 2} hence: x/2  2 =  2 , x = 0
f) 0.01 ^{x} = 100 = 10^{ 2} = (1 / 10^{ 2}) = (1 / 0.01) = 0.01 ^{ 1} , which gives 0.01 ^{x} = 0.01 ^{ 1}, hence x = 1
g) 2^{2x}  6(2^{x}) =  8
Note that 2^{2x} = (2^{x})^{ 2}.
Let u = 2^{x} and write u^{2} = (2^{x})^{ 2} = 2^{2x}
We now substitute 2^{x} by u and 2^{2x} by u^{2} in the given equation and rewrite the equation in terms of u only and in standard form as follows
u ^{2}  6 u + 8 = 0
Solve the above quadratic equation by factoring:
(u  2)(u  4) = 0
solutions in u: u = 2 and u = 4
We now solve for x using the substitution made above:
u = 2 = 2^{x} , x = 1
u = 4 = 2^{x} = 2^{2} , x = 2
More References and links
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