Applying Combined Exponents Rules: Factorization

To solve this exercise, first we note that 25 is the result of a power and we reduce it to a common base of 5.

$\sqrt{25}=5$$25=5^2$Now, we go back to the initial exercise and solve by adding the powers according to the formula:

$a^n\times a^m=a^{n+m}$

$5^4\times25=5^4\times5^2=5^{4+2}=5^6$

First, let's note that 27 is a power of the number 3:

$27=3^3$Using this fact gives us a situation where in the fraction's numerator and denominator we get terms with identical bases, let's apply this to the problem:

$\frac{27}{3^8}=\frac{3^3}{3^8}$Now let's recall the law of exponents for division between terms without identical bases:

$\frac{a^m}{a^n}=a^{m-n}$ Let's apply this law to the last expression we got:

$\frac{3^3}{3^8}=3^{3-8}=3^{-5}$where in the first stage we applied the above law and in the second stage we simplified the expression we got in the exponent,

Let's summarize the solution steps, we got:

$\frac{27}{3^8}=\frac{3^3}{3^8}=3^{-5}$Therefore the correct answer is answer D.

First, we recognize that 81 is a power of the number 3, which means that:

$3^4=81$We replace in the problem:

$\frac{81}{3^2}=\frac{3^4}{3^2}$Keep in mind that the numerator and denominator of the fraction have terms with the same base, therefore we use the property of powers to divide between terms with the same base:

$\frac{b^m}{b^n}=b^{m-n}$We apply it in the problem:

$\frac{3^4}{3^2}=3^{4-2}=3^2$$$Therefore, the correct answer is option b.

We use the formula:

$a^0=1$

$\frac{9\times3}{8^0}=\frac{9\times3}{1}=9\times3$

We know that:

$9=3^2$

Therefore, we obtain:

$3^2\times3=3^2\times3^1$

We use the formula:

$a^m\times a^n=a^{m+n}$

$3^2\times3^1=3^{2+1}=3^3$

We use the formula:

$(\frac{a}{b})^n=\frac{a^n}{b^n}$

$(\frac{2}{6})^3=(\frac{2}{2\times3})^3$

We simplify:

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

$\frac{1\times1\times1}{3\times3\times3}=\frac{1}{27}$

First we'll use the fact that raising any number to the power of 0 gives the result 1, mathematically:

$X^0=1$

We'll apply this to both the numerator and denominator of the fraction in the problem:

$\frac{4^0\cdot6^7}{36^4\cdot9^0}=\frac{1\cdot6^7}{36^4\cdot1}=\frac{6^7}{36^4}$

Next we'll note that -36 is a power of the number 6:

$36=6^2$

And we'll use this fact in the denominator to get expressions with identical bases in both the numerator and denominator:

$\frac{6^7}{36^4}=\frac{6^7}{(6^2)^4}$

Now we'll recall the power rule for power of a power to simplify the expression in the denominator:

$(a^m)^n=a^{m\cdot n}$

And we'll also recall the power rule for division between terms with identical bases:

$\frac{a^m}{a^n}=a^{m-n}$

We'll apply these two rules to the expression we got above:

$\frac{6^7}{(6^2)^4}=\frac{6^7}{6^{2\cdot4}}=\frac{6^7}{6^8}=6^{7-8}=6^{-1}$

Where in the first stage we applied the first rule we mentioned earlier - the power of a power rule and simplified the expression in the exponent of the denominator term, then in the next stage we applied the second power rule mentioned before - the division rule for terms with identical bases, and again simplified the expression in the resulting exponent,

Finally we'll use the power rule for negative exponents:

$a^{-n}=\frac{1}{a^n}$

And we'll apply it to the expression we got:

$6^{-1}=\frac{1}{6}$

Let's summarize everything we did, we got that:

$\frac{4^0\cdot6^7}{36^4\cdot9^0}=\frac{1}{6}$

Therefore the correct answer is A.

First, let's note that 4 is a power of 2:

$4=2^2$therefore we can perform a conversion to a common base for all terms in the problem,

Let's apply this:

$\frac{2}{4^{-2}}=\frac{2}{(2^2)^{-2}}$Next, we'll use the power law for power of power:

$(a^m)^n=a^{m\cdot n}$and we'll apply this law to the denominator term we got in the last step:

$\frac{2}{(2^2)^{-2}}=\frac{2}{2^{2\cdot(-2)}}=\frac{2}{2^{-4}}$where in the first step we applied the above law to the denominator and in the second step we simplified the expression we got,

Next, we'll use the power law for division between terms with identical bases:

$\frac{a^m}{a^n}=a^{m-n}$and we'll apply this law to the last expression we got:

$\frac{2}{2^{-4}}=2^{1-(-4)}=2^{1+4}=2^5$

Therefore the correct answer is answer B.