A question in powers

Question

I am a student in the second secondary so I had a question in math about powers like 4 power what gives you half in mathematical way I mean if there is a rule or a theory which can help me if I fall to a something like that

Answer 1

You are looking for $x$ such that $$4^x = \frac12$$ For this, using logarithms, as already said in comments and answers, makes the problem simple. So, let us try something else.

Multiply both sides by $2$ and get $$2 \times 4^x=1$$ Now, take into account that $4=2 ^2$, so $4^x=2^{2x}$. So, now we write $$2 \times 2^{2x}=1$$ that is to say $$2^{2x+1}=1$$ But you know that whatever could be $a$, $a^0=1$. Then this implies for your problem $2x+1=0$ and then $x=-\frac 12$.

Answer 2

Yes, it's called the logarithm function : http://en.wikipedia.org/wiki/Logarithm

But this topic may be too advanced

Answer 3

What you are asking about is the solution of:

$$4^x = \frac12$$

To solve equations where the unknown is an exponent, the usual method is to apply the logarithm to both sides. This is because of the special property of logarithm: $\log(a^b) = b\log(a)$. Now:

$$4^x = \frac12 \implies \log(4^x) = \log(\frac12) \implies x\log(4) = \log(\frac12) \implies x = \frac{\log(\frac12)}{\log(4)}$$

Though $\log(4) = \log(2^2) = 2\log(2)$; and $\log(\frac12) = \log(2^{-1}) = -\log(2)$

Thus:

$$x = \frac{-\log(2)}{2\log(2)} = -\frac12$$

Answer 4

Or \begin{align} 4^x &= \dfrac 12 \\ (2^2)^x &= 2^{-1}\\ 2^{2x} &= 2^{-1}\\ 2x &= -1 \quad (\text{If } a^b = a^c \text{, then }b = c.)\\ x &= -\dfrac 12 \end{align}