In this document, I have 5 exercises involving exponents and their five corresponding solutions. I am trying to find out why these are the solutions for these equations, as I cannot seem to get the same results as the provided solutions. The exercises ask to simplify the following equations. Can anyone assist in explaining how to solve for these five equations?
Thank you!
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On
$1)$ Use the law of exponents ie $a^{-n}=\frac1{a^n}$
so $(-3)^4\div3^5=\frac{(-1)^43^4}{3^5}=3^{4-5}=3^{-1}$
$2)$ $\dfrac{2^{-1}+2^{-2}}{2^{-1}\cdot2^{-2}}=\frac{2^{-1}}{2^{-3}}+\frac{2^{-2}}{2^{-3}}= 2^2+2=6$
$3)$ note that $(a\cdot b)^c = a^c\cdot b^c$
$\dfrac{81^4\cdot4^9}{6^{16}}= \dfrac{3^{4\cdot4}\cdot2^{2\cdot9}}{3^{16}\cdot2^{16}}=\dfrac{3^{16}\cdot2^{18}}{3^{16}\cdot2^{16}}= 2^2 =4$
$4)$ remember that $(-1)^{even}= 1$ and $(-1)^{odd}=-1$
$-(-1)^2-(-1^5)-(-1)^7-(-1)^{100} = -(1)-(-1)-(-1)-(1) =-1+1+1-1 = 0$
$5)$ $a^0 =1$ so,
$\dfrac{(5^{-1}+5^{-2})^0}{5^{1}+5^{-2}}= \dfrac{1}{\frac15+\frac1{25}}=\frac{1}{\frac{5+1}{25}}=\dfrac{25}6$
Welcome to Math.SE Christian. It is always better if you trype out questions using mathjax: link to the guide on how to do that here. I'll answer your question anyway this time, but please bear this in mind in future.
When we work with powers, there are two critical rules to remember. $$a^n\cdot a^m=a^{n+m}[I]$$ $$\frac{a^n}{a^m}=a^{n-m} [II]$$
For the first question, note that $(-k)^n\equiv k^n$ IF $n$ is even. It is in this case, so you use rule $[II]$ for $$\frac{3^4}{3^5}=3^{4-5}=3^{-1}$$
For the second question: Note that $k^{-n}=\frac{1}{k^n}$, thus $2^{-1}=\frac 12, 2^{-2}=\frac{1}{2^2}=\frac 14$. Thus the numerator is: $$\frac 12 + \frac 14=\frac 34$$ and the denominator is: $$\frac 12 \cdot \frac 14 =\frac 18$$ To evaluate$\frac{(\frac 34)}{(\frac 18)}$, note this is equivalent to $\frac 34 \cdot \frac 81$ and go from there.
For your third question, some new rules come into play: $$m^xn^x=(mn)^x [III]$$ and $${(m^n)}^x=m^{nx} [IV]$$
Using these two, you can simplify your equation to: $$\frac{3^{16}\cdot2^{18}}{2^{16}\cdot 3^{16}}=2^2=4$$
For Question $4$, note $(-1)^n$ is $-1$ if $n$ is odd, and $1$ if $n$ is even. So your equation becomes: $$-(1)-(-1)-(-1)-(1)=-1+1+1-1=0$$ For the fifth question, note anything (except $0$) to the power $0$ is $1$.
Thus you have: $$\frac{1}{5^{-1}+5^{-2}}$$ Using the $5^{-n}=\frac{1}{5^n}$ rule, we can then find this is $$\frac{1}{(\frac{6}{25})}=\frac{25}{6}$$