Question

How do you simplify \[{10^a}{.10^b}{.10^c}\] ?

Answer 1

Hint: For positive real numbers, if the bases are the same, powers get added in case of multiplication, and subtracted in case of division. So using this concept we will get the result.

Complete step-by-step solution:
\[{10^a}{.10^b}{.10^c}\]
We can expand the above equation for more clarity and we will get
\[\overbrace {10 \times 10 \times ..... \times 10}^a \times \overbrace {10 \times 10 \times ..... \times 10}^b \times \overbrace {10 \times 10 \times ..... \times 10}^c\]
As it is quite visible that in the first term, \[10\] gets multiplied to\[10\] , \[a\] times. This whole term is again in multiplication with the second term, where \[10\] , is in multiplication with itself, \[b\] times. Again these two terms are in multiplication with \[10\] , which is also in multiplication with itself, \[c\] times. Here,\[a\] , \[b\] and \[c\] are variables and their values can be anything like positive or negative integers, fractions, or even zero.
Therefore,
\[ \Rightarrow \overbrace {10 \times 10 \times .. \times 10}^{a + b + c\,times}\]
The above expression can be simply written as
\[ \Rightarrow {10^{a + b + c}}\]

Hence, the result of \[{10^a}{.10^b}{.10^c}\] is \[{10^{a + b + c}}\]

Note: When a number is multiplied itself, it results in the exponential form of the number as \[x \times x = {x^2}\] , here, \[x\] is called the base and \[2\] is called the power or exponent.
Here, the bases are the same and are in multiplication with each other. Therefore, all the bases raised to whatever power, the powers will get added up.
Had the bases been the same and all the terms were in division, all the powers would get subtracted.
An important rule in case of integers bases except zero , having power zero is that the value in such cases is always . This means that any positive integer (except zero), any fraction when raised to a power of zero will always return one as the value. Mathematically, this can be written as
\[{x^0} = 1\]
Where is \[x\] any number mentioned above.