SOLUTION: Will someone please help me on these two problems. I was absent and have a test on them later today. 1. Use log5 2~0.4307 and log5 3~0.6826 to approximate the value of log5 12.

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Question 167591: Will someone please help me on these two problems. I was absent and have a test on them later today.
1. Use log5 2~0.4307 and log5 3~0.6826 to approximate the value of log5 12.
2. Express log6 19 in terms of common logarithms. Then approximate its value to four decimal places.
I must show my work. Can you please help me through these
Thanks

Answer by nerdybill(7384)   (Show Source): You can put this solution on YOUR website!
1. Use log5 2~0.4307 and log5 3~0.6826 to approximate the value of log5 12.
.
So, instead of
log5(12)
we can write:
log5(2*2*3)
Applying log rule below:
logb(mn) = logb(m) + logb(n)
.
we get:
log5(2*2*3) = log5(2) + log5(2) + log5(3)
Now, we can substitute in the provided values:
log5(2*2*3) = 0.4307 + 0.4307 + 0.6826
log5(2*2*3) = 1.544
.
2. Express log6 19 in terms of common logarithms. Then approximate its value to four decimal places.
.
For this, we'll need to apply the change-of-base formula:
logb(x) = log(x)/log(b)
where "log" without a base indicates base 10.
.
So, now we can rewrite:
log6(19) = log(19)/log(6)
At this point, you'll need your calculator:
log6(19) = 1.27875/0.77815
log6(19) = 1.6433


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