Question

(1/3)^3+a=
(1/27)^3b
If 9^a + 1 = 81^b +2 and
Find the values of a and b.​

Answer 1

Answer:

I don't know sorry bro

Step-by-step explanation:

Mark as brainlist

Answer 2

Step-by-step explanation:

$1. ({ \frac{1}{3} })^{3 + a} = ({ \frac{1}{27} })^{3b}$

$= > ({ \frac{1}{3} })^{3 + a} = ({ \frac{1}{3} )}^{3 (3b)}$

$= > ({ \frac{1}{3} )}^{3 + a} = ({ \frac{1}{3} })^{9b}$

$as \: it \: has \: same \: base \: so \: we \: have$

$3 + a = 9b$

$a = 9b - 3$

$a = 3(3b - 1)$

$now \: we \: have \: to \: solve \: {9}^{a + 1} = {81}^{b + 2}$

${9}^{a} + 1 = ({({9})^{2} })^{b + 2}$

${9}^{a + 1} = {9}^{2b + 4}$

$same \: base \: to \: cut \: so \: we \: have$

$a + 1 = 2b + 4$

$a + 1 - 4 = 2b$

$a - 3 = 2b$

$we \: \: have \: a = 3(3b - 1)$

$3(3b - 1) - 3 = 2b$

$9b - 3 - 3 = 2b$

$9b - 6 = 2b$

$9b - 2b = 6$

$7b = 6$

$b = \frac{6}{7}$

$now \: put \: the \: value \: of \: b \: in \: the \: value \: of \: a = 3(3b - 1)$

$a = 3(3( \frac{6}{7}) - 1)$

$a = 3( \frac{18}{7} - 1)$

$a = 3( \frac{18 - 7}{7} )$

$a = 3 \times \frac{11}{7}$

$a = \frac{33}{7}$

.•. so the value of a and b is 33/7 and 6/7