Ask your own question, for FREE!
Algebra 17 Online
OpenStudy (anonymous):

Use log10 (5) ≈ 0.6990 and log10 (7) ≈ 0.8451 to approximate the value of log10 (245) .

ganeshie8 (ganeshie8):

hint : 245 = 5 x 49

OpenStudy (anonymous):

I got ≈ 2.699

ganeshie8 (ganeshie8):

close, check again...

OpenStudy (anonymous):

Got the same answer :/

ganeshie8 (ganeshie8):

\(\large \log_{10} 245\) \(\large \log_{10} 5 * 49\) \(\large \log_{10} 5 + \log_{10} 49\) \(\large \log_{10} 5 + \log_{10} 7^2\) \(\large \log_{10} 5 + 2 * \log_{10} 7\) \(\large 0.6990 + 2*0.8451 \) \(\large 2.3892\)

OpenStudy (valpey):

log(A*B) = log(A) + log(B)

OpenStudy (anonymous):

Oh ok thanks...

ganeshie8 (ganeshie8):

np :)

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!