Question

How do you solve
`3^(x) = (1)/(9)`

Answer 1

x=-2

basically i did inverse method; opposite of divide is times so i did 9x1 which is 9, which leaves you with 3^x=9, then you do the rest which is -2 :)

Answer 2

x = -2

Explanation:

You want to solve 3^x = 1/9.

Solution

Assuming you recognize that 9 = 3·3 = 3², you can make use of the rules of exponents to rewrite 1/9 and match exponents:

`3^x=(1)/(9)=(1)/(3^2)=3^(-2)`

Matching exponents of 3 tells you ...

x = -2

Alternate solution

Exponential equations can often be solved using logarithms.

x·log(3) = log(1/9)

x = log(1/9)/log(3) = -2

Or you can rewrite the logarithm using the rules for that:

x = (log(1) -log(9))/log(3) = (0 -log(3²))/log(3)

= (0 -2·log(3))/log(3)

x = -2

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Additional comment

The relevant rule of exponents is ...

1/a^b = a^-b

The relevant rules of logarithms are ...

log(a/b) = log(a) -log(b)

log(a^b) = b·log(a)

log(1) = 0

Matching exponents of 3 is equivalent to taking the logarithm, base 3.

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