Question

(1/3)^x=18 can you compute this

Answer 1

We have the following equation,

$(\frac{1}{3})^x=18$

By taking logarithm base 3 to both sides, we have

$\log _3(\frac{1}{3})^x=\log _318$

By the powers rule for logarithms, we get

$x\cdot\log _3(\frac{1}{3})^{}=\log _318$

Now, by the quotient rule, we have that

$\begin{gathered} \log _3(\frac{1}{3})=\log _3(1)-\log _3(3) \\ \log _3(\frac{1}{3})=-\log _3(3) \end{gathered}$

because base 3 logarithm of 1 is zero. Then, we have

$-x\cdot\log _3(3)^{}=\log _318$

But

$\log _3(3)=1$

then, we obtain

$-x=\log _318$

By multiplying both sides by -1, we have

$x=-\log _318$

Finally, since

$\log _318=2.6309$

The answer is x= - 2.6309