TutorMe homepage

SIGN IN

Start Free Trial

Tenti T.

Tutor for 3 years

Tutor Satisfaction Guarantee

Applied Mathematics

TutorMe

Question:

Solve the following expression for x: $log_2(x) = 3$

Tenti T.

Answer:

To solve this expression we need to use the definition of logarithms which states that if: $$log_{a}(b) = c$$ then it is also true that $$a^c = b$$ so using the above definition with a = 2 , b = x and c = 3 we write $$log_{2}(x) = 3$$ then it is also true that $$2^3 = x$$ $$x = 2^3$$ $$x = 8$$

Calculus

TutorMe

Question:

Find the first derivative of the following expression $$f(x) = 2x^{3} $$

Tenti T.

Answer:

so we re write the expression as $$f'(x) = (2x^{3})'$$ the constant goes out of the derivative : $$f'(x) =2 (x^{3})'$$ now using the formula $$(x^{n} )'= n x^{n-1}$$ where n = 3 : $$f'(x) =2(3)(x^{3-1})'$$ This is our answer : $$f'(x) = 6x^{2}$$

Algebra

TutorMe

Question:

Solve the following systems of equations : 3x+5y = 7 3x + 4y = 1

Tenti T.

Answer:

by subtracting the first equation from the second we have : $$3x-3x+5y - 4y = 7 - 1$$ $$y = 6$$ now we substitute the value of y we found in the first equation and we get : $$3x + 5(6) = 7$$ $$3x = 7-30$$ $$3x = -23$$ $$x= -\dfrac{23}{3}$$

Send a message explaining your

needs and Tenti will reply soon.

needs and Tenti will reply soon.

Contact Tenti

Ready now? Request a lesson.

Start Session

FAQs

What is a lesson?

A lesson is virtual lesson space on our platform where you and a tutor can communicate.
You'll have the option to communicate using video/audio as well as text chat.
You can also upload documents, edit papers in real time and use our cutting-edge virtual whiteboard.

How do I begin a lesson?

If the tutor is currently online, you can click the "Start Session" button above.
If they are offline, you can always send them a message to schedule a lesson.

Who are TutorMe tutors?

Many of our tutors are current college students or recent graduates of top-tier universities
like MIT, Harvard and USC.
TutorMe has thousands of top-quality tutors available to work with you.

Made in California

© 2019 TutorMe.com, Inc.