How do you multiply #m^ { - 7/ 6} \cdot m ^ { 1/ 4}#?

Question

Ethan Adkins · Accepted Answer

#m^(-11/12)# Using the #color(blue)"law of exponents"# #color(orange)"Reminder " color(red)(bar(ul(|color(white)(2/2)color(black)(a^mxxa^n=a^(m+n))color(white)(2/2)|)))# #rArrm^(-7/6)xxm^(1/4)=m^(-7/6+1/4)# #"Now " -7/6+1/4=-28/24+6/24=-22/24=-11/12# #rArrm^(-7/6+1/4)=m^(-11/12)#

Kennedy Adams · Answer

#1/m^(11/12)#

Recall the product rule for exponents:

#color(blue)(bar(ul(|color(white)(a/a)a^m*a^n=a^(m+n)color(white)(a/a)|)))#

When you are multiplying two powers with the same base, you add their exponent values together.

Applying the rule to the given question,

#m^(-7/6)*m^(1/4)#

The expression becomes #m# to the power of #-7/6+1/4#.

#=m^(-7/6+1/4)#

Since the fractions being added together do not have a common denominator, rewrite each fraction so that each one has the same denominator.

#=m^(-14/12+3/12)#

Evaluating,

#=m^(-11/12)#

However, expressions with negative exponents are usually simplified so that it only contains positive exponents.

Recall the negative exponent rule:

#color(blue)(bar(ul(|color(white)(a/a)a^-m=1/a^mcolor(white)(a/a)|)))#

Hence, #m^(-11/12)# becomes

#=color(green)( bar (ul ( | color(white)(a/a) color(black)(1/m^(11/12)) color(white)(a/a) | )))#

Genesis Allen · Answer

undefined To multiply $m^{-\frac{7}{6}} \cdot m^{\frac{1}{4}}$, you add the exponents because they have the same base, which is $m$.

So, $$m^{-\frac{7}{6}} \cdot m^{\frac{1}{4}} = m^{-\frac{7}{6} + \frac{1}{4}}.$$

To add the exponents, you need a common denominator. The common denominator for $6$ and $4$ is $12$.

$$= m^{-\frac{7}{6} + \frac{3}{12}}.$$

Then, you convert $\frac{7}{6}$ to an equivalent fraction with denominator $12$.

$$\frac{7}{6} = \frac{7 \times 2}{6 \times 2} = \frac{14}{12}.$$

So, $$= m^{-\frac{14}{12} + \frac{3}{12}}.$$

Now, add the fractions.

$$\frac{-14}{12} + \frac{3}{12} = -\frac{11}{12}.$$

Thus, $$m^{-\frac{7}{6}} \cdot m^{\frac{1}{4}} = m^{-\frac{11}{12}}.$$