C_TSCM62_66 SAP Certification questions and answers
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C_TSCM62_66 SAP Certification questions and answers
Logarithm and Exponential Questions with Answers and Solutions - Grade 12 Grade 12 questions on Logarithm and exponential with answers and solutions are presented. 1. Solve the equation (1/2) 2x + 1 = 1
2. Solve x ym = y x3 for m.
3. Given: log 8(5) = b. Express log 4(10) in terms of b.
25. Note that b 4 logbx = x4 The given equation may be written as: 2x x 4 = 486 x = 2431/5 = 3
26. 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.
27. 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
28. Given: 9 x - 3x - 8 = 0 Note that: 9 x = (3x)2 Equation may be written as: (3 x)2 - 3x - 8 = 0 Let y = 3 x and rewite 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 3x = ( 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
29. Given: 4 x - 2 = 3x + 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 * 42) / ln (4/3) = ln (3 4 * 24) / ln (4/3) = 4 ln(6) / ln(4/3)
30. 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 = 24 = 16
Solve Logarithmic Equations - Detailed Solutions Solve logarithmic equations including some challenging questions. Detailed solutions are presented. The logarithmic equations in examples 4, 5, 6 and 7 involve logarithms with different bases and are therefore challenging. Example 1: Solve the logarithmic equation
log2(x - 1) = 5. Solution to example 1
Rewrite the logarithm as an exponential using the definition. x - 1 = 25
Solve the above equation for x. x = 33
check: Left Side of equation log2(x - 1) = log 2(33 - 1) = log 2(25) = 5 Right Side of equation = 5
conclusion: The solution to the above equation is x = 33
Example 2: Solve the logarithmic equation
log5(x - 2) + log 5(x + 2) = 1. Solution to example 2
Use the product rule to the expression in the right side. log5(x - 2)(x + 2) = 1
Rewrite the logarithm as an exponential (definition). (x - 2)(x + 2) = 5 1
Which can be simplified as. x2 = 9
Solve for x. x = 3 and x = -3
check: 1st solution x = 3 Left Side of equation: log 5(3 - 2) + log 5(3 + 2) = log 51 + log5(3 + 2) = log 55 = 1 Right Side of Equation 2nd solution x = -3 Left Side of equation: log 5(-3 - 2) + log 5(-3 + 2) = log 5(-5) + log5(-1) log5(-5) and log 5(-1) are both undefined and therefore x = -3 is not a solution. conclusion: The solutions to the given equation is x = 3
Example 3: Solve the logarithmic equation
log3(x - 2) + log 3(x - 4) = log 3(2x^2 + 139) - 1. Solution to example 3
We first replace 1 in the equation by log 3(3) and rewrite the equation as follows. log3(x - 2) + log 3(x - 4) = log 3(2x^2 + 139) - log 3(3)
We now use the product and quotient rules of the logarithm to rewrite the equation as follows. log3[ (x - 2)(x - 4) ] = log 3[ (2x^2 + 139) / 3 ]
Which gives the algebraic equation (x - 2)(x - 4) = (2x^2 + 139) / 3
Mutliply all terms by 3 and simplify 3(x - 2)(x - 4) = (2x^2 + 139)
Solve the above quadratic equation to obtain
x = -5 and x = 23
check: 1) x = - 5 cannot be a solution to the given equation as it would make the argument of the logarithmic functions on the right negative. 2) x = 23 Right Side of equation: log3(23 - 2) + log 3(23 - 4) = log 3(21*19) = log 3(399) Left Side of equation: log3(2(23)^2 + 139) - 1 = log 3(1197) - log 3(3) = log3(1197 / 3) = log 3(399)
conclusion: The solution to the above equation is x = 23
Example 4: Solve the logarithmic equation
log4(x + 1) + log 16(x + 1) = log 4(8). Solution to example 4
We first note that 2 logarithms in the given equation have base 4 and one has base 16. We first use the change of base formula to write that log16(x + 1) = log 4(x + 1) / log 4(16) = log4(x + 1) / 2 = log 4(x + 1)1/2
We now write the given equation as follows.
log4(x + 1) + log 4(x + 1) 1/2 = log4(8)
We use the product rule to write. log4(x + 1)(x + 1) 1/2 = log4(8)
Which gives (x + 1)(x + 1) 1/2 = (8)
which can be written as (x + 1)3/2 = (8)
Solve for x to obtain. x=3
check: Left Side of equation: log4(3 + 1) + log 16(3 + 1) = 1 + 1/2 = 3/2
Right Side of equation: log4(8) = log4(43/2) = 3/2 conclusion: The solution to the above equation is x = 3
Example 5: Solve the logarithmic equation
log2(x - 4) + log sqrt(2)(x3 - 2) + log0.5(x - 4) = 20. Solution to example 5
We first use the change of base formula to write. logsqrt(2)(x3 - 2) = log 2(x3 - 2) / log2(sqrt(2)) = 2log 2(x3 - 2)
We also use the change of base formula to write. log0.5(x - 4) = -log 2(x - 4)
Substitute into the equation and simplify the given equation. 2 log2(x3 - 2) = 20
rewrite as. log2(x3 - 2) = 10
which gives x3 - 2 = 210
Solve the above equation for x. x = cube_root (1026)
Example 6: Solve the logarithmic equation
ln(x + 6) + log(x + 6) = 4. Solution to example 6
Use the change of base formula to rewrite log(x + 6) as log(x + 6) = ln(x + 6) / ln(10)
and substitute in the given equation ln(x + 6) + ln(x + 6) / ln(10) = 4
