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 ym = y x3 for m.
-
Given: log8(5) = b. Express log4(10) in terms of b.
-
Simplify without calculator: log6(216) + [ log(42) - log(6) ] / log(49)
-
Simplify without calculator: ((3-1 - 9-1) / 6)1/3
-
Express (logxa)(logab) as a single logarithm.
-
Find a so that the graph of y = logax passes through the point (e , 2).
-
Find constant A such that log3 x = A log5x , for all x > 0.
-
Solve for x the equation log [ log (2 + log2(x + 1)) ] = 0
-
Solve for x the equation 2 x b4 logbx = 486
-
Solve for x the equation ln (x - 1) + ln (2x - 1) = 2 ln (x + 1)
-
Find the x intercept of the graph of y = 2 log( √(x - 1) - 2)
-
Solve for x the equation 9x - 3x - 8 = 0
-
Solve for x the equation 4x - 2 = 3x + 4
- If logx(1 / 8) = -3 / 4, than what is x?
Solutions to the Above Problems
-
Rewrite equation as (1/2)2x + 1 = (1/2)0
Leads to 2x + 1 = 0
Solve for x: x = -1/2
-
Divide all terms by x y and rewrite equation as: ym - 1 = x2
Take ln of both sides (m - 1) ln y = 2 ln x
Solve for m: m = 1 + 2 ln(x) / ln(y)
-
Use log rule of product: log4(10) = log4(2) + log4(5)
log4(2) = log4(41/2) = 1/2
Use change of base formula to write: log4(5) = log8(5) / log8(4) = b / (2/3) , since log8(4) = 2/3
log4(10) = log4(2) + log4(5) = (1 + 3b) / 2
-
log6(216) + [ log(42) - log(6) ] / log(49)
= log6(63) + log(42/6) / log(72)
= 3 + log(7) /2 log(7) = 3 + 1/2 = 7/2
-
((3-1 - 9-1) / 6)1/3
= ((1/3 - 1/9) / 6)1/3
= ((6 / 27) / 6)1/3 = 1/3
-
Use change of base formula: (logxa)(logab)
= logxa (logxb / logxa) = logxb
-
2 = logae
a2 = e
ln(a2) = ln e
2 ln a = 1
a = e1/2
-
Use change of base formula using ln to rewrite the given equation as follows
ln (x) / ln(3) = A ln(x) / ln(5)
A = ln(5) / ln(3)
-
Rewrite given equation as: log [ log (2 + log2(x + 1)) ] = log (1) , since log(1) = 0.
log (2 + log2(x + 1)) = 1
2 + log2(x + 1) = 10
log2(x + 1) = 8
x + 1 = 28
x = 28 - 1
-
Note that b4 logbx = x4
The given equation may be written as: 2x x4 = 486
2 x5 = 486
x = 2431/5 = 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) = 32
x - 1 = 9
x = 10
-
Given: 9x - 3x - 8 = 0
Note that: 9x = (3x)2
Equation may be written as: (3x)2 - 3x - 8 = 0
Let y = 3x and rewrite equation with y: y2 - y - 8 = 0
Solve for y: y = ( 1 + √(33) ) / 2 and ( 1 - √(33) ) / 2
Since y = 3x, the only acceptable solution is y = ( 1 + √(33) ) / 2
3x = ( 1 + √(33) ) / 2
Use ln on both sides: ln 3x = ln [ ( 1 + √(33) ) / 2]
Simplify and solve: x = ln [ ( 1 + √(33) ) / 2] / ln 3
-
Given: 4x - 2 = 3x + 4
Take ln of both sides: ln ( 4x - 2 ) = ln ( 3x + 4 )
Simplify: (x - 2) ln 4 = (x + 4) ln 3
Expand: x ln 4 - 2 ln 4 = x ln 3 + 4 ln 3
Group like terms: x ln 4 - x ln 3 = 4 ln 3 + 2 ln 4
Solve for x: x = ( 4 ln 3 + 2 ln 4 ) / (ln 4 - ln 3) = ln (34 * 42) / ln (4/3) = ln (34 * 24) / ln (4/3)
= 4 ln(6) / ln(4/3)
-
Rewrite the given equation using exponential form: x- 3 / 4 = 1 / 8
Raise both sides of the above equation to the power -4 / 3: (x- 3 / 4)- 4 / 3 = (1 / 8)- 4 / 3
simplify: x = 84 / 3 = 24 = 16
More on Logarithm and Exponential Questions with Answers and Solutions - Grade 11
High School Maths (Grades 10, 11 and 12) - Free Questions and Problems With Answers
Middle School Maths (Grades 6, 7, 8, 9) - Free Questions and Problems With Answers
Primary Maths (Grades 4 and 5) with Free Questions and Problems With Answers
Home Page