Help with Precalculus

Question

3.2 Logarithmic functions

write each exponential equation in logarithmic form

1.) a² + 2 =7 --------------> answer: log_a5 =2

2.) in 2 = x --------------> answer: e^x =2

Provide a video on how you were able to get that answer and steps

Evaluate each expression without using a calculator

1.) log₂ 1/8 --------------> answer: -3

2.) log₃ √27 --------------> answer: 3/2

Provide a video on how you were able to get that answer and step

Evaluate each expression

1.) log₆(6⁷) --------------> answer: 7

2.) 3^log₃⁵ --------------> answer: 5

Provide a video on how you were able to get that answer and step

find the domain of each function

1.) f(x) = log₂(x+1) --------------> answer: (-1, ∞)

2.) f(x) = log₃ √x-1 --------------> answer: (1, ∞)

Provide a video on how you were able to get that answer and step

Answer 1

1.

a^2 + 2 = 7

Subtract 2 from both sides of the above equation.

a^2 = 5

Take the log base a of both sides.

log_a (a^2) = log_a (5) and recall that log a is the inverse function of the exponent function a^x, so log a (a^x) = x.

2 = log a (5)

or, using the symmetric property of equality,

log a (5) = 2

2.

ln (2) = x

Take both sides to the power e and recall that e^(ln(x)) = x, since these are inverse functions.

2 = e^x

Or

e^x = 2

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1.

log 2 (1/8) = log 2 (1) - log 2 (8) = 0 - log 2 (2^3) = -3

2.

log 3 (√27) = log 3 [27^(1/2)] = (1/2) log 3 (27)

= (1/2) log 3 (3^3) = (1/2) × (3) = 3/2

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1.

log 6 (6^7) = 7 × log 6 (6) = 7× 1 = 7

2.

3^(log 3 (5) ) = 5, since the exponentation of a log base 3 to the base 3 inverts the function and just leaves 5.

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1.

The log function inputs of any acceptable base must be greater than 0.

So, for

log 2 ( x + 1 )

The input x+1 > 0 is required.

Subtracting 1 from both sides,

x > -1 and that is the domain for f(x).

In interval notation, this is ( -1 , ∞ ).

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2.

Iog 3 ( √ (x - 1 ) )

The input to any acceptable log function must be greater than 0.

In addition, the input to the square root function must be greater than or equal to zero if you are using the real number system.

Thus x - 1 ≥ 0 if and only if x ≥ 1, which is easy to show by adding 1 to both sides.

The log functions requires the inputs to be positive, so we further restrict the input to be x > 1.

Thus f(x) must have a domain of x > 1.

In interval notation, this is ( 1 , ∞ ).

Answer 2

This video explains the questions: https://rumble.com/v3id5uc-log-properties-1.html