Efficient Logarithmic Calculations for 0.3048 without a Calculator

[RChristenk]

Homework Statement

Given ##\log2=0.3010300##, ##\log3=0.4771213##, ##\log7=0.8450980##, find ##\log0.3048##

Relevant Equations

Logarithm rules

##0.3048=\dfrac{3048}{10000}=\dfrac{2^3\cdot3\cdot127}{10^4}##

##\log0.3048=\log(\dfrac{2^3\cdot3\cdot127}{10^4})##

##\Rightarrow 3\log2+\log3+\log127-4\log10##

I don't have the value for ##\log127##, and this problem is to be solved without a calculator. All the logarithms are base ##10##. I'm not sure how else to factorize ##0.3048##. Is there another way entirely or some artificial artifice I can use here? Thanks.

 

[PeroK]

I'm not sure I'd worry about this problem. I can't see a solution.

 

[FactChecker]

As an approximation, can you use ##2^8 = 128##? CORRECTION: ##2^7=128##
##2^3*3*128 = 3072##.

 

[Hill]

As an approximation, can you use ##2^8 = 128##?
##2^3*3*128 = 3072##.

But then, what's ##\log 7## for?

##126##?

Both?

 

[FactChecker]

But then, what's ##\log 7## for?

##126##?

Both?

Good point! So both approximations are doable with the information given. I don't know which one was expected or better. It seems like 126 was expected.

 

[FactChecker]

Good point! So both approximations are doable with the information given. I don't know which one was expected or better. It seems like 126 was expected.

Or you could do both approximations and interpolate.

 

[Hill]

Good point! So both approximations are doable with the information given. I don't know which one was expected or better. It seems like 126 was expected.

Maybe, the midpoint between ##\log 126## and ##\log 128##.

 

[fresh_42]

As an approximation, can you use ##2^8 = 128##?
##2^3*3*128 = 3072##.

##2^8=256.## It's ##\log 127 \approx 7\log 2.##

 

[FactChecker]

##2^8=256.## It's ##\log 127 \approx 7\log 2.##

Thanks! I stand corrected and will note that in the post.