Question

By what number should ${3^9}$ be multiplied so as to get 3.

Answer 1

Hint:
We can assume x to be the required number. Then we can multiply the given number with x and equate it to 3. Then we can solve for x. Then we can obtain the required number by using properties of exponents.

Complete step by step solution:
Let the required number that gives 3 when multiplied with ${3^9}$ be x.
Now we can write it as an equation.
 $ \Rightarrow {3^9} \times x = 3$
Now we can divide both sides with ${3^9}$ . So, we will get,
 $ \Rightarrow x = \dfrac{3}{{{3^9}}}$
We can write the numerator 3 as 3 raised to the power 1. So, we will get,
 $ \Rightarrow x = \dfrac{{{3^1}}}{{{3^9}}}$
We know that by properties of exponents, $\dfrac{{{a^x}}}{{{a^y}}} = {a^{x - y}}$ . On applying this condition, we get,
 $ \Rightarrow x = {3^{1 - 9}}$
On simplifying the power, we get,
 $ \Rightarrow x = {3^{ - 8}}$

So, the required number is ${3^{ - 8}}$.

Note:
Alternate solution is given by,
As the given number is a power of 3, the required number also will be a power of 3.
Let the required number that gives 3 when multiplied with ${3^9}$ be ${3^x}$.
Now we can write it as an equation.
 $ \Rightarrow {3^9} \times {3^x} = 3$
Now we can take the logarithm on both sides. So, we get,
 $ \Rightarrow \log \left( {{3^9} \times {3^x}} \right) = \log 3$
We know that the log of the products is equal to the sum of the logarithms. It is given by the equation $\log ab = \log a + \log b$. Then the equation will become
 $ \Rightarrow \log \left( {{3^9}} \right) + \log \left( {{3^x}} \right) = \log 3$
We know that $\log {a^b} = b\log a$. So, the equation will become,
 \[ \Rightarrow 9\log \left( 3 \right) + x\log \left( 3 \right) = \log 3\]
Now we can divide throughout with $\log 3$. So, the equation will become,
 \[ \Rightarrow 9 + x = 1\]
Now we can solve for x.
 \[ \Rightarrow x = 1 - 9\]
On further simplification, we get,
 \[ \Rightarrow x = - 8\]
So, the required number is ${3^{ - 8}}$.