question

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

\(\large \color{black}{\begin{align} \text{find }\ \log_{10}3\ \text{without calculator }\ \hspace{.33em}\\~\\ \end{align}}\)
It really depends on how accurate you'd like it to be... We can use the power series of \(\ln(1+x)\) to make a decent approximation. \[\ln(1+x)=-\sum_{n=1}^\infty\frac{(-x)^n}{n}=x-\frac{x^2}{2}+\frac{x^3}{3}-\cdots\] Change the base to the natural one, then use the series above: \[\log_{10}3=\frac{\ln3}{\ln10}=\frac{\ln(1+2)}{\ln(1+9)}\] At least that's how I would approach it if I didn't have the luxury of using a calculator.

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

i need 2 decimal places
how long do need to write the first series for 2 decimal places
approximation correct upto 2 decimal places

Not the answer you are looking for?

Search for more explanations.

Ask your own question