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.

Simplify without calculator: log_{6}(216) + [ log(42)  log(6) ] / log(49)

Simplify without calculator: ((3^{1}  9^{1}) / 6)^{1/3}

Express (log_{x}a)(log_{a}b) as a single logarithm.

Find a so that the graph of y = log_{a}x passes through the point (e , 2).

Find constant A such that log_{3} x = A log_{5}x , for all x > 0.

Solve for x the equation log [ log (2 + log_{2}(x + 1)) ] = 0

Solve for x the equation 2 x b^{4 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 9^{x}  3^{x}  8 = 0

Solve for x the equation 4^{x  2} = 3^{x + 4}
 If log_{x}(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: y^{m  1} = x^{2}
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: log_{4}(10) = log_{4}(2) + log_{4}(5)
log_{4}(2) = log_{4}(4^{1/2}) = 1/2
Use change of base formula to write: log_{4}(5) = log_{8}(5) / log_{8}(4) = b / (2/3) , since log_{8}(4) = 2/3
log_{4}(10) = log_{4}(2) + log_{4}(5) = (1 + 3b) / 2

log_{6}(216) + [ log(42)  log(6) ] / log(49)
= log_{6}(6^{3}) + log(42/6) / log(7^{2})
= 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: (log_{x}a)(log_{a}b)
= log_{x}a (log_{x}b / log_{x}a) = log_{x}b

2 = log_{a}e
a^{2} = e
ln(a^{2}) = ln e
2 ln a = 1
a = e^{1/2}

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 + log_{2}(x + 1)) ] = log (1) , since log(1) = 0.
log (2 + log_{2}(x + 1)) = 1
2 + log_{2}(x + 1) = 10
log_{2}(x + 1) = 8
x + 1 = 2^{8}
x = 2^{8}  1

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

Given: 4^{x  2} = 3^{x + 4}
Take ln of both sides: ln ( 4^{x  2} ) = ln ( 3^{x + 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 (3^{4} * 4^{2}) / ln (4/3) = ln (3^{4} * 2^{4}) / 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 = 8^{4 / 3} = 2^{4} = 16
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