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Mathematics for Computing












































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Let a, x and y be real numbers, u and v be positive numbers, and a > 0 with a ≠ 1. Which of the following is/are correct?

(a) log_a (2) = 1
(b) log_a (u/v) = log_a (u-v)
(c) if a^x = y, then x = log_a (y)
(d) log_a (9) = 2log_a (3)
(e) log_a (uv) = log_a (u) + log_a (v)

Comments for Mathematics for Computing

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Jun 10, 2011
Logarithms
by: Staff

The question:

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As closely as I can tell, this is what you submitted:

Let a, x and y be real numbers, u and v be positive numbers, and a > 0 with a ≠ 1. Which of the following is/are correct?

(a) log_a (2) = 1
(b) log_a (u/v) = log_a (u-v)
(c) if a^x = y, then x = log_a (y)
(d) log_a (9) = 2log_a (3)
(e) log_a (uv) = log_a (u) + log_a (v)



The answer:


This might help to put logarithms in perspective:

Multiplication is a shortcut for addition: 3 + 3 + 3 + 3 = 4 * 3, four times three

Exponents are a shortcut for multiplication: 3 * 3 * 3 * 3 = 3⁴, three to the 4th power (the exponent is 4)

Logarithms are a shortcut for exponents: log_3 (81) = log_3 (3⁴) = 4, the exponent is 4. The log of 81 using a base of 3 is 4 because the exponent is 4. 81 = 3⁴.





(a) log_a (2) = 1

Correct for a=2 (a base of 2). log_2 (2¹) = 1



(b) log_a (u/v) = log_a (u-v)

Incorrect

Remember that logarithms stand for exponents. The rules of exponents apply to logarithms.


For example, 81/3 = (3⁴)/ (3¹)

= (3⁴⁻¹)

= 3³

log_3 (3³) = 3, the exponent is 3


Therefore,

log_3 (81/3)

= log_3 (3⁴/3¹)

>>>>>>>>>>>>> = log_3 (3⁴) - log_3 (3¹)

= 4 - 1

= 3, the exponent is 3



The correct answer to part (b) is: log_a (u/v) = log_a (u) – log_a (v)



(c) if a^x = y, then x = log_a (y)

Correct.

Start with the original equation:

a^x = y

take the log_a of each side of the equation

log_a (a^x) = log_a (y)

x = log_a (y)




(d) log_a (9) = 2log_a (3)

Correct.

log_a (9) = log_a (3*3)

log_a (9) = log_a (3) + log_a (3)

log_a (9) = 2*log_a (3)



(e) log_a (uv) = log_a (u) + log_a (v)

Correct.






Thanks for writing.

Staff
www.solving-math-problems.com


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