How do you solve 3^ { 3r - 1} = 3^ { - 2r - 3}33r1=32r3?

2 Answers

In the given eqn, the bases are equal, so equate powers on both sides.

Explanation:

3r-1=-2r-33r1=2r3

5r=-25r=2

r= -2/5r=25

Apr 16, 2017

r=-2/5r=25

Explanation:

Since the color(blue)"bases on both sides of the equation are 3"bases on both sides of the equation are 3 we can equate the exponents.

rArr3r-1=-2r-33r1=2r3

rArr5r=-25r=2

rArrr=-2/5r=25

color(blue)"As a check"As a check

3^((-6/5-5/5))=3^(-11/5)3(6555)=3115

"and " 3^((4/5-15/5))=3^(-11/5)and 3(45155)=3115

rArrr=-2/5" is the solution"r=25 is the solution