Logarithm and Exponential Questions with Answers and Solutions  Grade 12
Grade 12 questions on Logarithm and exponential with answers and solutions are presented.
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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( sqrt(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
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( sqrt(x  1)  2)
Divide both sides by 2: log( sqrt(x  1)  2) = 0
Use the fact that log(1)= 0: sqrt(x  1)  2 = 1
Rewrite as: sqrt(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 + sqrt(33) ) / 2 and ( 1  sqrt(33) ) / 2
Since y = 3^{x}, the only acceptable solution is y = ( 1 + sqrt(33) ) / 2
3^{x} = ( 1 + sqrt(33) ) / 2
Use ln on both sides: ln 3^{x} = ln [ ( 1 + sqrt(33) ) / 2]
Simplify and solve: x = ln [ ( 1 + sqrt(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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