How do I find the number whose common logarithm is 2.6025?

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How do I find the number whose common logarithm is 2.6025?

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Answer 1 · 2015-07-14T01:49:31

Can use ${log}_{10} (2) ≅ 0.30103$ $log_10(2) ~= 0.30103$ to find approximate value $400$ $400$ then multiply by an approximation for $10^{0.00044} ≅ 1.001$ $10^0.00044 ~= 1.001$ to get $400.4$ $400.4$ .

If you have a calculator then just $10^{2.6025} ≅ 400.4055$ $10^(2.6025) ~= 400.4055$

Explanation:

$10^{2.6025} = 10^{2} \cdot 10^{0.60206} \cdot 10^{0.00044}$ $10^(2.6025) = 10^2*10^0.60206*10^0.00044$

$≅ 100 \cdot 10^{2 {log}_{10} (2)} \cdot 10^{0.00044}$ $~=100*10^(2log_10(2))*10^0.00044$

$= 100 \cdot 2^{2} \cdot 10^{0.00044}$ $=100*2^2*10^0.00044$

$= 400 \cdot 10^{0.00044}$ $=400*10^0.00044$

Now

${1.001}^{1000} = {(1 + 0.001)}^{1000}$ $1.001^1000 = (1+0.001)^1000$

$=1+((1000),(1))0.001+((1000),(2))0.001^2+...$

$=1+1+0.4995+0.166167+0.04141712475+...$

somewhere between $2.7$ and $3$

$log_10(3) ~= 0.4771$ (one of those useful numbers to memorise)

$log_10(2.7) = log_10(27/10) = log_10(3^3/10) = 3log_10(3)-1$

$~= 0.4313$

So $0.4313 < 1000 log_10(1.001) < 0.4771$

$0.0004313 < log_10(1.001) < 0.0004771$

So $1.001$ is a fairly good approximation for $10^0.00044$

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How do I find the number whose common logarithm is 2.6025?

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