If log 2 ≈ 0.301 and log 3 ≈ 0.477, what is the approximate value of log 72? How to do this
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If log 2 ≈ 0.301 and log 3 ≈ 0.477, what is the approximate value of log 72? How to do this

[From: ] [author: ] [Date: 12-04-22] [Hit: ]
Hope this helped.72 = 2³3² (i.e.--> you have a multiplication of two numbers.........
A. 0.051
B. 0.778
C. 0.861
D. 1.857

Don't tell me to put in calculator.

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You don't need a calculator. You need to know the rules of logarithms and you need to know how to factor numbers

log(a^b) = b * log(a)
log(a * b) = log(a) + log(b)
log(a/b) = log(a) - log(b)

log(72) =>
log(8 * 9) =>
log(8) + log(9) =>
log(2^3) + log(3^2) =>
3 * log(2) + 2 * log(3) =>
3 * 0.301 + 2 * 0.477 =>
0.903 + 0.954 =>
1.857

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For this answer, no real calculations are required. Since the log function on a calculator is to base 10, the question would look like this:

10^x = 72

10 to the power of anything smaller than 1 would give you an answer that would be less than 10. There is only one answer above that has an exponent larger than 1. The answer is D.

The first part of the question is letting you know what the base of the log function is. (10^0.301=2 , 10^0.477=3)

Hope this helped.

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72 = 2*36 = 2 * 2 * 18 = 2 * 2 * 2 * 9 = 2 * 2 * 2 * 3 * 3
-->

72 = 2³3² (i.e. 8 * 9)

Therefore we can write:

log(72) = log(2³3²)
--> you have a multiplication of two numbers...separate using log(a * b) = log(a) + log(b)

log(2³3²) = log(2³) + log(3²)
--> now use log(a^b) = b * log(a)

log(2³) + log(3²) = 3 * log(2) + 2 * log(3)
--> now plug in your values for log(2) and log(3)

log(72) ~ 3 * .301 + 2 * .477
-->

.903 + .954 = 1.875, the answer is D)

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You need to use the properties of logs.

log(MN) = logM + logN
log(Aˣ) = x logA

log(72) = log(2³ * 3²) = log 2³ + log 3² = 3 log 2 + 2 log 3

--------> 3 * .301 + 2 * .477 = .903 +.954 = 1.857 ---> D

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log(72)
= log(9x8)
= log(9) + log(8)
= log(3²) + log(2³)
= 2log(3) + 3log(2)
≈ 2x0.477 + 3x0.301
= 0.954 + 0.903
= 1.857

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since 72 = 2*2*2*3*3
log (72) = log (2*2*2*3*3)
log (72) = log (2) + log (2) log (2) + log (3) + log (3)
log (72) = .301 + .301 +.301 +.477 +.477 = 1.857

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log 72 =
log 8*9 =
log8 + log9 =
log2^3 + log3^2 =
3log2 + 2log3 =
3(0.301) + 2(0.477) = 1.857
Answer : D

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D

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