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Five-Minute Check (over Lesson 7–5)
CCSS
Then/Now
New Vocabulary
Example 1: Find Common Logarithms
Example 2: Real-World Example: Solve Logarithmic Equations
Example 3: Solve Exponential Equations Using Logarithms
Example 4: Solve Exponential Inequalities Using Logarithms
Key Concept: Change of Base Formula
Example 5: Change of Base Formula
Over Lesson 7–5
A. 1.9864
B. 2.3885
C. 3.1547
D. 4
Use log3 4 ≈ 1.2619 and log3 8 ≈ 1.8928 to approximate the value of log3 32.
Over Lesson 7–5
A. –0.6309
B. 0.1577
C. 0.3155
D. 0.4732
Use log3 4 ≈ 1.2619 and log3 8 ≈ 1.8928 to approximate the value of log3 .
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Over Lesson 7–5
Which of the following equations is false?
A. log8 m5 = 5 log8 m
B. loga 6 – loga 3 = loga 2
C.
D. logb 2x = logb 2 + logb x
Content Standards
A.CED.1 Create equations and inequalities in one variable and use them to solve problems.
Mathematical Practices
4 Model with mathematics.
You simplified expressions and solved equations using properties of logarithms.
• Solve exponential equations and inequalities using common logarithms.
• Evaluate logarithmic expressions using the Change of Base Formula.
Find Common Logarithms
A. Use a calculator to evaluate log 6 to the nearest ten-thousandth.
Answer: about 0.7782
Keystrokes: ENTERLOG 6 .7781512504
Find Common Logarithms
B. Use a calculator to evaluate log 0.35 to the nearest ten-thousandth.
Answer: about –0.4559
Keystrokes: ENTERLOG .35 –.4559319556
Solve Logarithmic Equations
Original equation
JET ENGINES The loudness L, in decibels, of a
sound is where I is the intensity of
the sound and m is the minimum intensity of sound detectable by the human ear. The sound of a jet engine can reach a loudness of 125 decibels. How many times the minimum intensity of audible sound is this, if m is defined to be 1?
Solve Logarithmic Equations
Exponential form
Answer: The sound of a jet engine is approximately 3 × 1012 or 3 trillion times the minimum intensity of sound detectable by the human ear.
Use a calculator.I ≈ 3.162 × 1012
Replace L with 125 and m with 1.
Divide each side by 10 and simplify.
A. 1,585,000,000 times the minimum intensityB. 1,629,000,000 times the minimum intensityC. 1,912,000,000 times the minimum intensityD. 2,788,000,000 times the minimum intensity
DEMOLITION The loudness L, in decibels, of a
sound is where I is the intensity of the
sound and m is the minimum intensity of sound detectable by the human ear. Refer to Example 2. The sound of the demolition of an old building can reach a loudness of 92 decibels. How many times the minimum intensity of audible sound is this, if m is defined to be 1?
Solve Exponential Equations Using Logarithms
Solve 5x = 62. Round to the nearest ten-thousandth.
5x = 62 Original equation
log 5x = log 62Property of Equality for Logarithms
x log 5= log 62Power Property of Logarithms
Answer: about 2.5643
x ≈ 2.5643 Use a calculator.
Divide each side by log 5.
Solve Exponential Equations Using Logarithms
Check You can check this answer by using a calculator or by using estimation. Since 52 = 25 and 53 = 125, the value of x is between 2 and 3. Thus, 2.5643 is a reasonable solution.
A. x = 0.3878
B. x = 2.5713
C. x = 2.5789
D. x = 5.6667
What is the solution to the equation 3x = 17?
Solve Exponential Inequalities Using Logarithms
Solve 37x > 25x – 3. Round to the nearest ten-thousandth.
37x
> 25x – 3
Original inequality
log 37x
> log 25x – 3
Property of Inequality for Logarithmic Functions
7x log 3
> (5x – 3) log 2
Power Property of Logarithms
Solve Exponential Inequalities Using Logarithms
7x log 3
> 5x log 2 – 3 log 2
Distributive Property
7x log 3 – 5x log 2
> – 3 log 2
Subtract 5x log 2 from each side.
x(7 log 3 – 5 log 2) > –3 log 2
Distributive Property
x > –0.4922 Simplify.
Solve Exponential Inequalities Using Logarithms
Use a calculator.
Divide each side by 7 log 3 – 5 log 2.
Solve Exponential Inequalities Using Logarithms
Check: Test x = 0.
37x > 25x – 3 Original inequality
Answer: The solution set is {x | x > –0.4922}.
?37(0)> 25(0) – 3 Replace x with 0.?30 > 2–3 Simplify.
Negative Exponent Property
A. {x | x > –1.8233}
B. {x | x < 0.9538}
C. {x | x > –0.9538}
D. {x | x < –1.8233}
What is the solution to 53x < 10x – 2?
Change of Base Formula
Express log5 140 in terms of common logarithms. Then round to the nearest ten-thousandth.
Answer: The value of log5 140 is approximately 3.0704.
Use a calculator.
Change of Base Formula