Solutions to the Above Questions

1. 
   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
   
   
2. 
   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) x² = a     as a logarithm     2 = log _x (a)
   
   
3. 
   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 = 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
   
4. 
   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) log_x 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) log_5.6 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) log_x 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.
   
5. 
   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) 2^2x - 6(2^x) = - 8
   Note that 2^2x = (2^x) ².
   Let u = 2^x and write u² = (2^x) ² = 2^2x
   We now substitute 2^x by u and 2^2x 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 = 2^x , x = 1
   u = 4 = 2^x = 2² , x = 2