Question

Expand log (1125/32)​

Answer 1

Answer:

Step-by-step explanation:

It depends where you are starting from.

Suppose you know:

log(2)≈0.30102999566

log(3)≈0.47712125472

Then you can calculate a good approximation for log(1125)

1125=32⋅53=32⋅(102)3=32⋅10323

So:

log(1125)=log(32⋅10323)

=log(32)+log(103)−log(23)

=2log(3)+3log(10)−3log(2)

=2log(3)+3−3log(2)

≈2⋅0.47712125472+3−3⋅0.30102999566

≈3.0511525225

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Answer 2

Answer:

3log5 + 2log3 - 5log2

Solution:

We have,

$\log \dfrac{1125}{32}$

Using the logarithm identity,

$\log \dfrac{a}{b}=\log a-\log b$

we can rewrite the solution as log1125 - log 32

1125 = 9* 5^3.

9 = 3^2

now using $\log a^n=n\log a$

we can rewrite log1125 as 3log5 + 2log3

for log 32,

32 = 2^5

so we can rewrite it as 5log2

Now we add all those up and get 3log5 + 2log3 - 5log2