Question

Given the log2 =0.3010 and log3=0.4771 calculatewithout using mathematical tables or calculator the value of
a.log54
b.log2.4
c.log30
d.log540

Answer 1

a. log54 = log(3^3 * 2) = 3log3 + log2 = 3(0.4771) + 0.3010 = 1.7323
b. log2.4 = log2 + log(1.2) = 0.3010 + log(6/5) [logarithmic identity: loga.b = loga + logb] = 0.3010 + log6 - log5 - log2 [logarithmic identities: loga/b = loga - logb, loga^b = b*loga] = 0.3010 + 1.7782 - 0.6989 - 1 = 0.3803
c. log30 = log(3 * 10) = log3 + log10 = log3 + 1 [logarithmic identity: loga.b = loga + logb] = 0.4771 + 1 = 1.4771
d. log540 = log(3^3 * 2^2 * 5) = 3log3 + 2log2 + log5 = 3(0.4771) + 2(0.3010) + 0.6989 = 2.7323

Answer 2

log 2.4 = log 24 - log 10 = log 24 - 1
= log (2*2*2*3) - 1 = log 2 + log 2 + log 2 + log 3 - 1

Answer 3

That is correct, but note that the value of log24 that you used is not exact (it is approximately 1.3802), so the final answer will have some degree of approximation.

Answer 4

... but we are not given log 5 , just log 2 and log 3
so I used log (10/2) = log 10 - log 2 = 1 - log 2

Answer 5

You're right, my apologies for my oversight. You used the proper method to evaluate log 2.4.