Logarithm and Exponential Questions with Answers and Solutions - Grade 12
The concepts of logarithm and exponential are used throughout mathematics. Questions on Logarithm and exponential with solutions, at the bottom of the page, are presented with detailed explanations.
Solve the equation (1/2)^{2x + 1} = 1
Solve x y^{m} = y x^{3} for m.
Given: log_{8}(5) = b. Express log_{4}(10) in terms of b.
Note that b^{4 logbx} = x^{4}
The given equation may be written as: 2x x^{4} = 486
2 x^{5} = 486
x = 243^{1/5} = 3
Group terms and use power rule: ln (x - 1)(2x - 1) = ln (x + 1)^{2}
ln function is a one to one function, hence: (x - 1)(2x - 1) = (x + 1)^{2}
Solve the above quadratic function: x = 0 and x = 5
Only x = 5 is a valid solution to the equation given above since x = 0 is not in the domain of the expressions making the equations.
Solve: 0 = 2 log( √(x - 1) - 2)
Divide both sides by 2: log( √(x - 1) - 2) = 0
Use the fact that log(1)= 0: √(x - 1) - 2 = 1
Rewrite as: √(x - 1) = 3
Raise both sides to the power 2: (x - 1) = 3^{2}
x - 1 = 9
x = 10
Given: 9^{x} - 3^{x} - 8 = 0
Note that: 9^{x} = (3^{x})^{2}
Equation may be written as: (3^{x})^{2} - 3^{x} - 8 = 0
Let y = 3^{x} and rewrite equation with y: y^{2} - y - 8 = 0
Solve for y: y = ( 1 + √(33) ) / 2 and ( 1 - √(33) ) / 2
Since y = 3^{x}, the only acceptable solution is y = ( 1 + √(33) ) / 2
3^{x} = ( 1 + √(33) ) / 2
Use ln on both sides: ln 3^{x} = ln [ ( 1 + √(33) ) / 2]
Simplify and solve: x = ln [ ( 1 + √(33) ) / 2] / ln 3