# About tutor mathstutor494

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### Welcome to mathstutor494's "About Me" page!

mathstutor494(95)
About mathstutor494: Retired engineer. Volunteer math tutor for high and middle schools algebra, arithmetic, geometry and trigonometry.

#### What did mathstutor494 acccomplish on this site?

• mathstutor494 has no published lessons yet.
• mathstutor494 has no published solvers yet.
• mathstutor494 wrote 95 solutions (see them).

Comment from student: Thank you so much for helping me with my homework

Comment from student: thank you very much for helping me. i HIGHLY appreciate it. thank you very much(:

Comment from student: thnaks, sir for helping it was a really good shortcut for solving that problem &wish you a very happy new year.

Comment from student: Thank you so much for the help. It really is awesome that I could use your help.. Have a wonderful day!!! Thanks Again!!!!

Comment from student: thanks a lot sir

Comment from student: Thank you:)

Comment from student: THANKS A LOT.

Comment from student: (I thought it was erroneous, too. {It's 2log (subscript"b").} I think there is an error in signs). If the orig. EQ. is still unsolvable, I'm providing an alternate (below).I really appreciate your fast reply and your effort. Note: in below, the term "2-a" is NOT log b. (Solve for x): log b(x) = 2-a + log b (a^2*b^a)/b^2)

Comment from student: Thank you!

Comment from student: Thank you very much!

Comment from student: thank y0u so much .. thanks .. very well thank y0u ..

Comment from student: Thank you, mathstutor494! Your solution pretty much explains the logic behind the question.

Comment from student: thanks very much for ur answer

Comment from student: thank you very much ... :D

Comment from student: Thank You so much, and I will recommend you to everybody I know that needs help :-)!

Comment from student: thanks

Comment from student:

Comment from student: Thank you so much your help! It really means a lot. :)

Comment from student: Thank you for the help.

Comment from student: Thank you for the help.