Logarithm and Exponential Questions with Answers and Solutions

Grade 11 questions on Logarithm and exponential with answers and solutions 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: y = log b (x) ⇔ x = b y where the symbol ⇔ means "is equivalent to", y is the exponent, b is the base such that b > 0 , b ≠ 1 and x > 0 Numerical example: 2 = log 3 (9) ⇔ 9 = 3 2 2) One-to-one properties of logarithmic and exponential functions a) If b x = b y then x = y b) If logb x = logb y then x = y Change the given logarithmic expressions into exponential expressions: a) logx (a) = c b) logb (2x + 1) = 3 Change the given exponential expressions into logarithmic expressions: a) 3x = m b) x2 = a Evaluate , without calculator, he following logarithmic expressions: a) log2 16 b) log3 27 c) log2 (1/32) d) log25 5 e) Log √(10) f) logb 1 , with b > 0 and b ≠ 0 g) log0.1 10 Solve for x the following logarithmic equations: a) log2 x = 3 b) logx 8 = 3 c) log3 x = 1 d) log5.6 x = 0 e) log2 (3x + 1) = 4 f) log3 (1/(x + 1)) = 2 g) log4 ( (x + 1)/(2x - 1) ) = 0 h) Log ( 1/x + 1 ) = 2 i) logx 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) 22x - 6(2x) = - 8 Solutions to the Above Questions Solution Use the equivalent expressions : y = log b (x) ⇔ x = b y to write a) logx (a) = c     as an exponential     a = x c b) logb (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) 3x = m     as a logarithm     x = log 3 (m) b) x2 = 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 = log2 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 = log2 16 = 4 b) let y = log3 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 = log3 27 = 3 c) let y = log2 (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 = log2 (1/32) = -5 d) let y = log25 5 convert to exponential form: 25 y = 5 = √(25) = 25 1/2 which gives 25 y = 25 1/2 hence y = log25 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 = logb 1 , ( with b > 0 and b ≠ 0) ; convert to exponential form: b y = 1 = b 0, hence y = logb 1 = 0 g) let y = log0.1 10 ; convert to exponential form: 0.1 y = 10 = 1 / 0.1 = (0.1) -1 , hence y = log0.1 10 = -1 Solution Use the equivalent expressions : y = log b (x) ⇔ x = b y to solve for x the following logarithmic equations: a) log2 x = 3 ; convert to exponential form: x = 2 3 = 8 b) logx 8 = 3 ; convert to exponential form: 8 = x 3 , write 8 as 8 = 2 3 ; hence x = 2 c) log3 x = 1 ; convert to exponential form: x = 3 1 = 3 d) log5.6 x = 0 ; convert to exponential form: x = 5.6 0 = 1 e) log2 (3x + 1) = 4 ; convert to exponential form: 3x + 1 = 2 4 = 16 , solve for x: 3x + 1 = 16 , 3x = 15 , x = 5 f) log3 (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) log4 ( (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) logx 0.0001 = 4 ; convert to exponential form: x 4 = 0.0001 = 1/10000 = 1/104 = (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 = 42, which gives 4 2x + 1 = 42, 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) 22x - 6(2x) = - 8 Note that 22x = (2x) 2. Let u = 2x and write u2 = (2x) 2 = 22x We now substitute 2x by u and 22x by u2 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 = 2x , x = 1 u = 4 = 2x = 22 , x = 2

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