Question

Select all the correct answers.

Which expressions are equivalent to the given expression?

5 log2o+ log10 20 – log10 10

o logo(100x) + 1

log, (2.1)

log2:5)

o log10 (10r)

o log10 (2005) – 1

Answer 1

Answer:

$5 \log_{10}x+ \log_{10} 20 - \log_{10} 10 = \log_{10}(2x^5)$

$5 \log_{10}x+ \log_{10} 20 - \log_{10} 10 = \log_{10}(20x^5) - 1$

Step-by-step explanation:

Given

$5 \log_{10}x+ \log_{10} 20 - \log_{10} 10$

Required

Equivalent expression

Rewrite the expression as:

$5 \log_{10}x+ \log_{10} 20 - \log_{10} 10 = \log_{10}x^5+ \log_{10} 20 - \log_{10} 10$

Apply law of logarithm

$5 \log_{10}x+ \log_{10} 20 - \log_{10} 10 = \log_{10}(\frac{x^5*20}{10})$

$5 \log_{10}x+ \log_{10} 20 - \log_{10} 10 = \log_{10}(x^5*2)$

$5 \log_{10}x+ \log_{10} 20 - \log_{10} 10 = \log_{10}(2x^5)$

Another possible equivalent expression is:

$5 \log_{10}x+ \log_{10} 20 - \log_{10} 10 = \log_{10}x^5+ \log_{10} 20 - \log_{10} 10$

Apply law of logarithm

$5 \log_{10}x+ \log_{10} 20 - \log_{10} 10 = \log_{10}(x^5*20) - \log_{10} 10$

$5 \log_{10}x+ \log_{10} 20 - \log_{10} 10 = \log_{10}(20x^5) - \log_{10} 10$

$\log_{10} 10 = 1$

So:

$5 \log_{10}x+ \log_{10} 20 - \log_{10} 10 = \log_{10}(20x^5) - 1$

Answer 2

The expression that is equivalent to the expression 5log₁₀x + log₁₀20 - log₁₀10 will be D. log₁₀(20x⁵) - 1.

How to find the equivalent expression?

From the information given, the expression given is 5log₁₀x + log₁₀20 - log₁₀10. The first thing to do is to apply the addition rule. = log₁₀(20x⁵) - log₁₀10.

Note that log₁₀10 = 1

Therefore, log₁₀(20x⁵) - log1010 will be log₁₀(20x⁵) - 1.

In conclusion, the correct option is D.

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