10 to the power of 3.5: $10^{3.5}$

So $10^3 = 10\times 10\times 10 = 1000$, this is really easy to understand.

But what about: $\,10^{3.5}\,?\,$ My logic would suggest this was

$10\times 10 \times 10\times 5 = 5000,\;$ but the calculator says it's 3162.27...

Can someone illustrate how the calculator calculates the power of when the number is with decimals?

Please keep in mind that I'm a newbie to mathematics!

• write 3.5 as $\frac {7}{2}$ so 7 is normal power and 1/2 is square root May 11, 2013 at 22:17
• I say leave this question open. This question is actually excellent. The OP has given a good well defined question and I believe that a good answer might actually help the OP with his/her understanding of the topic. May 12, 2013 at 18:58
• I just want to point out the flaw of logic in your argument. What did you expect $1^{0.5}$ to be ? May 12, 2013 at 19:50

With $10^{0.5}$ you have to do a "half multiplication", not a multiplication by half. What this "half multiplication" is doesn't follow from any universal law, but only by extending in a coherent way the rules valid for integer exponents.

Since, for integer $m$ and $n$ you have $$a^{m+n}=a^m\cdot a^n$$ you can derive also that $$(a^m)^n=a^{mn}$$ (just repeat the multiplications and count the factors). So, what $a^{3.5}$ should mean? Well, a possible choice comes from doing $$a^7=a^{3.5\cdot 2}\overset{*}{=}(a^{3.5})^2$$ where the equals sign marked with $*$ is where we apply an extension to the rule above.

Thus one can try defining $$a^{3.5}=\sqrt{a^7}.$$

This is how actually exponentiation to a rational is defined: $$a^{\frac{p}{q}}=\sqrt[q]{a^p}$$ and it can be shown that the rules

$$a^{x+y}=a^x\cdot a^y,\qquad (a^x)^y=a^{xy}$$ continue to hold for all rational numbers $x$ and $y$ (and positive $a$).

• How about irrational numbers? Such as a^(1/9) ? May 12, 2013 at 8:43
• @Rafi: Where's the irrational number? May 12, 2013 at 8:45
• @RafiKamal That requires much more work. You can see this answer for a quick definition. However $1/9$ is not an irrational number. May 12, 2013 at 8:46

$$10^{3.5}=10^3\cdot 10^{.5}=10^3 \cdot 10^{\frac{1}{2}}=10^3 \sqrt{10}$$

Half power doesn't mean half of the number, it means square root.

• While very true, it doesn't help with the visualization part of the question: if $10^3 = 10\cdot 10 \cdot 10$, then $10^{3.5} =\ 10\cdot 10 \cdot 10 \cdot\ ?$ May 12, 2013 at 1:09
• @Virtlink how about $10^3 = 10^1 * 10^1 * 10^1, so 10^3.5 = 10^1 * 10^1 * 10^1 * 10^0.5? (sorry, can't figure out how to format that with tex) May 12, 2013 at 1:45 • @Virtlink$10^{3.5}=10\cdot 10 \cdot 10 \cdot \sqrt{10}$. May 12, 2013 at 2:50$10^{3.5}$is equal to$10*10*10*10^{0.5}$. So you just need to know what$10^{0.5}$is. One of the property of exponent is this.$(10^x)^y = 10^{xy}$, i.e. the power of a power, is just the exponents multiplied. So if$(10^{0.5})^2=x^2=10^1$when$x=10^{0.5}$then we just solve for$x^2 = 10$. I'm not sure if you learned about square roots, yet, but$x=\sqrt{10}$. Since$\sqrt{10}$is approximately equal to 3.16227, your calculator gives you$10^{3.5} = 3162.27...$. • Thank you so much, so many good answers!!, no i havent learned about square roots yet, but im on it now. May 11, 2013 at 22:32 Your mistake is computing$10^\frac{6}{2} \cdot \frac{10}{2}$instead of$10^\frac{6}{2} \cdot 10^\frac{1}{2}$. Although 5 is halfway to ten when working with addition, it is past halfway to ten when working with multiplication.$10^{3.5}$is equivalent to$10^\frac{7}{2}$. You can perform the '7' and 'over 2' parts of the exponentiation separately. That is to say:$10^\frac{7}{2} = (10^7)^\frac{1}{2}$. That's just$\sqrt{10^7}$, which can be reduced to$10^3 \sqrt{10}$, which is approximately$3163.3$. N.S. is exactly correct;$10^.5 = \sqrt{10}$. For a small amount of reasoning behind it. By exponent laws, we know$10^x \times 10^y = 10^{x+y}$.$10^.5 \times 10^.5 = 10^1 = 10$. Since we want to figure out what$10^.5$is, let's sub in$x$for it.$x \times x = 10 \rightarrow \sqrt{x^2}=\sqrt{10} \to x = \sqrt{10}$There was a question about irrational numbers. For example, how would$10^{\pi}$be calculated? The trick is to use$ln(x)$and$e^{x}$which are inverse functions, that is,$e^{ln(x)} = x$If you then use$x = a^{n}$, you'll get$a^{n} = e^{ln(a^{n})} = e^{n ln(a)}$or$10^{\pi} = e^{\pi ln(10)}$Both$ln(x)$and$e^{x}\$ can be easily calculated.