# If 1/81 = 3^x, then what is x???Please solve in both the ways i.e., using algebra and logarithm...

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`1/81 = 3^x`

(I) Using algebraic method.

Express 81 with its prime factors.

`1/3^4 = 3^x`

Apply the negative exponent rule which which is `1/a^m=a^(-m)` .

`3^(-4)=3^x`

Since both sides of the equation has the same base 3, equate the exponents of each side equal to each other.

`-4=x`

**Hence, the solution is `x=-4` .**

(II) To bring down x, apply the power rule of logarithm which is

`log_b m^a = a log_bm` . So take the logarithm of both sides, with a base of 3.

`log_3 1/81 = log_3 3^x`

`log_3 1/81 = x log_3 3`

At the left side of the equation, apply the quotient rule of logarithm `log_b (m/n) = log_b m - log_b n` .

`log_3 1 - log_3 81 = x log_3 3`

Express 81 with its prime factors.

`log_3 1 - log_3 3^4 = x log_3 3`

`log_3 1 - 4log_3 3 = x log_3 3`

Note that a logarithm of 1 is always equal to zero ( `log_b 1 = 0` ).Also, if the base and argument are the same, then the logarithm is equal to one (`log_b b = 1` ).

`0 -4(1) = x(1)`

`-4=x`

**Hence, `x=-4` .**

Algebraic Method:

1/81 = 3^x

=> 81 = 3^(-x)

=> 3^4 = 3^(-x)

Therefore

- x = 4

=> x = - 4

Logarythmic Method

1/81 = 3^x

=> 3^4 = 3^(-x)

=> 4 * log(3) = - x log(3), The log is of base "e" or any other real

positive fixed base.

=> x = - 4

So the solution is

x = - 4