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.