SOLUTION: if log10^2=0.3010 and log10 ^7=0.8451.Evaluate "1" log10 ^0.2 "2" log10^35^2 without using tables or calculator .Evaluate non

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: if log10^2=0.3010 and log10 ^7=0.8451.Evaluate "1" log10 ^0.2 "2" log10^35^2 without using tables or calculator .Evaluate non      Log On


   

Question 1205369: if log10^2=0.3010 and log10 ^7=0.8451.Evaluate "1" log10 ^0.2 "2" log10^35^2 without using tables or calculator .Evaluate
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Found 3 solutions by math_tutor2020, Alan3354, ikleyn:
Answer by math_tutor2020(3817) About Me  (Show Source):
You can put this solution on YOUR website!

Problem 1

Each log shown below is base 10.
log%28%282%29%29+=+0.3010 (given)
log%28%287%29%29+=+0.8451 (given)


So,
log%28%280.2%29%29+=+log%28%282%2F10%29%29

log%28%280.2%29%29+=+log%28%282%29%29+-+log%28%2810%29%29 Use log rule log(A/B) = log(A)-log(B)

log%28%280.2%29%29+=+0.3010+-+1 Use the rule that log(10) = 1 where the log is base 10.

log%28%280.2%29%29+=+-0.699 This is approximate.

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Problem 2

Each log shown below is base 10.
log%28%2835%5E2%29%29+=+2%2Alog%28%2835%29%29 Use log rule log(A^B) = B*log(A) to pull down the exponent.

log%28%2835%5E2%29%29+=+2%2Alog%28%285%2A7%29%29 Factor 35 as 5*7.

log%28%2835%5E2%29%29+=+2%2A%28log%28%285%29%29%2Blog%28%287%29%29%29 Use log rule log(AB) = log(A)+log(B)

log%28%2835%5E2%29%29+=+2%2A%28log%28%2810%2F2%29%29%2Blog%28%287%29%29%29 Rewrite the 5 as 10/2.

log%28%2835%5E2%29%29+=+2%2A%28log%28%2810%29%29+-+log%28%282%29%29%2Blog%28%287%29%29%29 Use log rule log(A/B) = log(A)-log(B)

log%28%2835%5E2%29%29+=+2%2A%281+-+0.3010%2B0.8451%29 Plug in the given info; also log(10) = 1 when log is base 10.

log%28%2835%5E2%29%29+=+3.0882 approximately

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Another approach for problem 2

Each log shown below is base 10.
log%28%2835%5E2%29%29+=+2%2Alog%28%2835%29%29

log%28%2835%5E2%29%29+=+2%2Alog%28%285%2A7%29%29

log%28%2835%5E2%29%29+=+2%2Alog%28%28%2810%2F2%29%2A7%29%29

log%28%2835%5E2%29%29+=+2%2Alog%28%287%2F%282%2F10%29%29%29

log%28%2835%5E2%29%29+=+2%2A%28log%28%287%29%29+-+log%28%282%2F10%29%29%29

log%28%2835%5E2%29%29+=+2%2A%28log%28%287%29%29+-+log%28%280.2%29%29%29

log%28%2835%5E2%29%29+=+2%2A%280.8451+-+%28-0.699%29%29 Plug in log(7) = 0.8451, and use the result of problem 1.

log%28%2835%5E2%29%29+=+3.0882 approximately

Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
if log10^2=0.3010 and log10 ^7=0.8451.Evaluate "1" log10 ^0.2 "2" log10^35^2 without using tables or calculator
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log(2) = 0.301 ---> log(5) = 1-0.301 = 0.699
35 = 5*7 --> log(35) = log(5) + log(7) = 1.5441
log(35^2) = 2*log(35) = 3.0882

Answer by ikleyn(52810) About Me  (Show Source):
You can put this solution on YOUR website!
.

Writing is TERRIBLE in this post, and is only good to scare people around.

Normal wording, if you do not know on how to write it correctly, should be like THIS 

    "log 2 base 10 is 0.3010, . . . "   and so on . . .



Also, do not use "1" or "2" to denote the number of a problem.

Use the standard notations 1) and 2) for it, or (1) and (2).

Also, write each sub-problem in separate line.


Do not try to confuse readers - - - if it is not your special goal.