Given:
[tex](\frac{1}{9})^{a+1} = 81^{a+1}\cdot 27^{2-a}[/tex]
To find:
a=?
Solution:
[tex](\frac{1}{9})^{a+1} = 81^{a+1}\cdot 27^{2-a}[/tex]
taking log on both sides:
[tex]\log ((\frac{1}{9})^{a+1}) =\log( 81^{a+1}\cdot 27^{2-a})\\\\{(a+1)} (\frac{1}{9}) =\log( 81^{a+1}) + \log(27^{2-a})\\\\{(a+1)} (\frac{1}{9}) = (a+1)81 + (2-a)27\\\\\frac{1}{9}a +\frac{1}{9} = 81a+81 + 54-27a\\\\\frac{1}{9}a +\frac{1}{9} = 54a+135\\\\\frac{1}{9}a -54a= 135- \frac{1}{9}\\\\\frac{1- 486}{9}a =\frac{1215- 1}{9}\\\\a =\frac{1214}{9}\times \frac{9}{-485}\\\\a =\frac{1214}{9}\times \frac{9}{-485}[/tex]
by solving the value we get "- 4"
Learn more:
brainly.com/question/12724725
