# Find a if 9^3a = 57.

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### 3 Answers

We have to find given that 9^3a = 57.

Now 9^3a = 57

take the logarithm to the base 3 on both the sides

=> log(3) [9^3a] = log (3) 57

=> 3a* log (3) 9 = log (3) [ 19*3]

=> 3a * log (3) 3^2 = log (3) 19 + log (3) 3

=> 3a * 2 = log(3) 19 + 1

=> 6a = log(3) 19 + 1

=> a = [log(3) 19 + 1]/6

=> a = [ 2.680 + 1]/6

=> a = 3.680/6

=> a = 0.6133

**Therefore a is equal to 0.6133.**

9^3a = 57.

We take the logarithms of both sides:

=> log{9^(3a)} = log57.

=> 3a log 9 = log57, as log a^m = mloga.

=> a = log57/3log9.

=> a = 1.755875/3*0.954243.

Therefore a = 0.6134 nearly.

We notice that we don't have common bases. We'll use logarithms to determine the variable a.

We'll take logarithms both sides:

log3 (9^3a) = log3 57

We'll use power property of logarithms:

3a*log3 9 = log3 57

But log3 9 = log3 3^2, where log3 3 = 1

log3 3^2 = 2log3 3

log3 3^2 = 2

3a*2 = log3 57

6a = log3 3*19

We'll use product property of logarithms:

6a = log3 3 + log3 19

6a = 1 + log3 19

log3 19 = lg 19/lg 3

log3 19 = 2.70

We'll divide by 6:

a = 1/6 + 2.70/6

a = 0.16 + 0.45

**a = 0.61**