SOLUTION: (1) If(2/3)^m=72/243, find m. (2) find the coefficient of x^8 in (x^2+2y/x)^10

Question 757375: (1) If(2/3)^m=72/243, find m. (2) find the coefficient of x^8 in (x^2+2y/x)^10
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
(1) If(2/3)^m=72/243, find m.
---
Since the variable is in the exponent, take the log of both sides.
m*log(2/3) = log(72/243)
-------
m = [log(72/243)]/[log(2/3)] = 3
==============

(2) find the coefficient of x^8 in (x^2+2y/x)^10
-------
Form of each term of the expansion:
(x^2)^(10-a)(2y/x)^a = x^(20-2a)/x^a = x^(20-3a)
---
Solve for "a"
20-3a = 8
-3a = -12
a = 4
10-a = 6
-------
The term with a = 4 and 10-a = 6
10C6 (x^2)^6*(2y/x)^4 = 210[x^12 * 16y^4/x^4]
--------
= 210*16 x^8*y^4
----
= 3360 x^8y^4
=================
Cheers,
Stan H.
==================

RELATED QUESTIONS
Given: m<1=2x+4o m<2=2y+40 m<3=x+2y Find: m<1, m<2,... (answered by stanbon)

Find the value of x in the following equation (1/9)^m x (27)^-1=243 (answered by Alan3354,josgarithmetic)

6) if m<1=x+30, m<2=x-23, and m<3=2x-7, find x and the value of each numbered angle.... (answered by stanbon)

the sum of three consecutive integers is 243. find the smallest interger. here's the (answered by venugopalramana)

find the coefficient of x^2 in the expansion of {x -... (answered by greenestamps)

If m<2 = x/2 -10 and M<3= x/3 + 40, find x and... (answered by cleomenius)

Expand (1/2-2x)^5 up to the term in x^3. If the coefficient of x^2 in the expansion of... (answered by greenestamps)

if m (angle) 1=3x+10 and m(angle) 2=x+16 and c//d, find the value of... (answered by solver91311)

If m<1 = x. find: M<2 M<3 M<4 Trying to figure out how to find the measurement... (answered by jim_thompson5910)