All Solutions
Section 4-3: Working with Rational Exponents
{text{ Remember these rules}} hfill \
{text{ }}{a^{frac{n}{m}}} = sqrt[m]{{{a^n}}} hfill \
sqrt[m]{{{a^m}}} = a hfill \
hfill \
{text{a) }}{49^{frac{1}{2}}} = sqrt {49} = sqrt {{7^2}} = 7 hfill \
hfill \
{text{b) }}{100^{frac{1}{2}}} = sqrt {100} = sqrt {{{10}^2}} = 10 hfill \
hfill \
{text{c) }}{left( { – 125} right)^{frac{1}{3}}} = sqrt[3]{{ – 125}} = sqrt[3]{{{{left( { – 5} right)}^3}}} = – 5 hfill \
hfill \
{text{d) }}{16^{0.25}} = {16^{frac{1}{4}}} = sqrt[4]{{16}} = sqrt[4]{{{2^4}}} = 2 hfill \
hfill \
{text{e) 8}}{{text{1}}^{frac{1}{4}}} = sqrt[4]{{81}} = sqrt[4]{{{3^4}}} = 3 hfill \
hfill \
{text{f) }} – {left( {144} right)^{0.5}} = – {left( {144} right)^{frac{1}{2}}} = – sqrt {144} = – sqrt {{{12}^2}} = – 12 hfill \
hfill \
end{gathered} ]
b) $sqrt{100}=10$
c) $sqrt[3]{-125}=-5$
d) $sqrt[4]{16}=2$
e) $sqrt[4]{81}=3$
f) $-sqrt{144}=-12$
a) $sqrt[9]{512}=512^{1/9}=(2^9)^{1/9}=2$
b) $sqrt[3]{-27}=(-27)^{1/3}=[(-3)^3]^{1/3}=-3$
c) $sqrt[3]{27^2}=27^{2/3}=(3^3)^{2/3}=3^2=9$
d) $left(sqrt[3]{-216}right)^5=(-216)^{5/3}=[(-6)^3]^{5/3}=(-6)^5=-7776$
e) $sqrt[5]{dfrac{-32}{243}}=left(dfrac{-32}{243}right)^{1/5}=left[dfrac{(-2)^5}{3^5}right]^{1/5}=-dfrac{2}{3}$
f) $sqrt[4]{left(dfrac{16}{81}right)^{-1}}=left(dfrac{16}{81}right)^{-1/4}=left(dfrac{81}{16}right)^{1/4} =left(dfrac{3^4}{2^4}right)^{1/4}=dfrac{3}{2}$
b) $(-27)^{frac{1}{3}}=-3$
c) $27^{frac{2}{3}}=9$
d) $(-216)^{frac{1}{5}}=7776$
e) $left(dfrac{-32}{243}right)^{frac{1}{5}}=-dfrac{2}{3}$
f) $left(dfrac{16}{81}right)^{-frac{1}{4}}=dfrac{3}{2}$
$$
8^{frac{2}{3}}cdot 8^{frac{1}{3}}=8^{left(frac{2}{3}+frac{1}{3}right)}=8^{frac{3}{3}}=8^1
$$
$8^{2/3}div8^{1/3}=8^{frac{2}{3}-frac{1}{3}}=8^{1/3}$
$$
(-11)^2(-11)^{frac{3}{4}}=(-11)^{left(2+frac{3}{4} right)}=(-11)^{left(frac{8}{4}+frac{3}{4} right)}=(-11)^{frac{11}{4} }
$$
$$
left(7^{frac{5}{6}}right)^{-frac{6}{5}}=7^{left( frac{5}{6}times -frac{6}{5}right)}=7^{-1}
$$
$$
dfrac{9^{-1/5}}{9^{2/3}}=9^{frac{-1}{5}-frac{2}{3}}=9^{-frac{3}{15}-frac{10}{15}}=9^{-frac{13}{15}}
$$
$10^{-frac{4}{5}}left( 10^{frac{1}{15}}right)div10^{frac{2}{3}}$
$=10^{left(-frac{4}{5}+frac{1}{15}right)}div 10^{2/3}$
$=10^{left(-frac{12}{15}+frac{1}{15}right)}div 10^{2/3}$
$=10^{-frac{11}{15}}div 10^{frac{2}{3}}$
Remember that $a^mdiv a^n=a^{m-n}$
$=10^{frac{-11}{15}-frac{2}{3}}$
$$
=10^{frac{-11}{15}-frac{10}{15}}=10^{-frac{21}{15}}=10^{-frac{7times 3}{5times 3}}=10^{-frac{7}{5}}
$$
b) $8^{frac{1}{3}}$
c) $(-11)^{frac{11}{4}}$
d) $7^{-1}$
e) $9^{frac{-13}{15}}$
f) $10^{-frac{7}{5}}$
$sqrt{5}sqrt{5}=sqrt{5^2}=5$
$=dfrac{sqrt[3]{-16}}{sqrt[3]{2}}=sqrt[3]{dfrac{-16}{2}}=sqrt[3]{-8}$
Remember that $sqrt[m]{a^n}=a^{n/m}$
$$
=sqrt[3]{-8}=sqrt[3]{(-2)^3}=-2
$$
$$
dfrac{sqrt{28}sqrt{4}}{sqrt{7}}=sqrt{dfrac{28}{7}}sqrt{4}=sqrt{4}sqrt{4}=sqrt{4^2}=4
$$
$dfrac{sqrt[4]{18}(sqrt[4]{9})}{sqrt[4]{2}}=sqrt[4]{dfrac{18}{2}}(sqrt[4]{9})=sqrt[4]{9}sqrt[4]{9}=sqrt[4]{9^2}=9^{2/4}$
$$
=9^{1/2}=sqrt{9}=3
$$
b) $-2$
c) $4$
d) $3$
{text{a) Remember this rule, }}{a^{frac{n}{m}}} = sqrt[m]{{{a^n}}} = {left( {sqrt[m]{a}} right)^n} hfill \
{49^{frac{1}{2}}} + {16^{frac{1}{2}}} hfill \
= sqrt {49} + sqrt {16} hfill \
= sqrt {{7^2}} + sqrt {{4^2}} hfill \
= 7 + 4 = 11 hfill \
hfill \
{text{b) Remember this rule, }}{a^{frac{n}{m}}} = sqrt[m]{{{a^n}}} = {left( {sqrt[m]{a}} right)^n} hfill \
{text{2}}{{text{7}}^{frac{2}{3}}} – {81^{frac{3}{4}}} = sqrt[3]{{{{27}^2}}} – sqrt[4]{{{{81}^4}}} hfill \
= {left( {sqrt[3]{{27}}} right)^2} – {left( {sqrt[4]{{81}}} right)^3} hfill \
= {left( {sqrt[3]{{{3^3}}}} right)^2} – {left( {sqrt[4]{{{3^4}}}} right)^3} hfill \
= {3^2} – {3^3} hfill \
= 9 – 27 hfill \
= – 18 hfill \
hfill \
{text{c) Remember this rule, }}{a^{frac{n}{m}}} = sqrt[m]{{{a^n}}} = {left( a right)^n} hfill \
{text{1}}{{text{6}}^{frac{3}{4}}} + {text{1}}{{text{6}}^{frac{3}{4}}} – {81^{ – frac{1}{4}}} = sqrt[4]{{{{16}^3}}} + sqrt[4]{{{{16}^3}}} – {81^{ – 1/4}} hfill \
= {left( {sqrt[4]{{16}}} right)^3} + {left( {sqrt[4]{{16}}} right)^3} – frac{1}{{sqrt[4]{{{{81}^{ – 1}}}}}} hfill \
= {left( {sqrt[4]{{{2^4}}}} right)^3} + {left( {sqrt[4]{{{2^4}}}} right)^3} – {left( {{3^4}} right)^{ – 1/4}} hfill \
= {2^3} + {2^3} – {3^{ – 1}} hfill \
= 8 + 8 – frac{1}{3} hfill \
= 16 – frac{1}{3} hfill \
= frac{{48}}{3} – frac{1}{3} = frac{{47}}{3} hfill \
hfill \
hfill \
end{gathered} ]
{text{d) Change to exponential form with a base of 2}} hfill \
hfill \
{text{12}}{{text{8}}^{ – frac{5}{7}}} – {16^{0.75}} = {left( {{2^7}} right)^{ – frac{5}{7}}} – {left( {{2^4}} right)^{frac{3}{4}}} hfill \
= {2^{ – 5}} – {2^3} hfill \
= frac{1}{{{2^5}}} – {2^3} hfill \
= frac{1}{{32}} – 8 hfill \
= frac{1}{{32}} – frac{{256}}{{32}} hfill \
= – frac{{255}}{{32}} hfill \
hfill \
{text{e) Change the exponent to fraction form}}{text{.}} hfill \
{text{1}}{{text{6}}^{frac{3}{2}}} + {16^{ – 0.5}} + 8 – {27^{frac{2}{3}}} = {text{1}}{{text{6}}^{frac{3}{2}}} + {16^{ – frac{1}{2}}} + 8 – {27^{frac{2}{3}}} hfill \
{text{Remember this rule, }}{a^{frac{n}{m}}} = sqrt[m]{{{x^n}}} = {left( {sqrt[m]{x}} right)^n} hfill \
= {left( {{2^4}} right)^{3/2}} + {left( {{4^2}} right)^{ – 1/2}} + 8 – {left( {{3^3}} right)^{2/3}} hfill \
= {2^6} + {4^{ – 1}} + 8 – {3^2} hfill \
= {2^6} + frac{1}{4} + 8 – 9 hfill \
= 64 + frac{1}{4} – 1 hfill \
= 63 + frac{1}{4} = frac{{252}}{4} + frac{1}{4} hfill \
= frac{{253}}{4} hfill \
hfill \
{text{f) 8}}{{text{1}}^{frac{1}{2}}} + sqrt[3]{8} – {32^{frac{4}{5}}} + {16^{frac{3}{4}}} = sqrt {81} + sqrt[3]{8} – {left( {sqrt[5]{{32}}} right)^4} + {left( {sqrt[4]{{16}}} right)^3} hfill \
= sqrt {{9^2}} + sqrt[3]{{{2^3}}} – {left( {sqrt[5]{{{2^5}}}} right)^4} + {left( {sqrt[4]{{{2^4}}}} right)^3} hfill \
= 9 + 2 – {2^4} + {2^3} hfill \
= 11 – 16 + 8 hfill \
= 3 hfill \
end{gathered} ]
b) $-18$
c) $dfrac{47}{3}$
d) $-dfrac{255}{32}$
e) $dfrac{253}{4}$
f) $3$
{text{a) Remember that }}{a^m} times {a^n} = {a^{m + n}} hfill \
{text{ }}{{text{4}}^{frac{1}{5}}}left( {{4^{0.3}}} right) = {4^{frac{1}{5} + frac{3}{{10}}}} hfill \
hfill \
= {4^{frac{2}{{10}} + frac{3}{{10}}}} hfill \
= {4^{frac{5}{{10}}}} = {4^{frac{1}{2}}} hfill \
= sqrt 4 = 2 hfill \
end{gathered} ]
{text{b) Remember that }}{a^m} times {a^n} = {a^{m + n}} hfill \
{text{10}}{{text{0}}^{0.2}}left( {{{100}^{frac{{ – 7}}{{10}}}}} right) = {100^{frac{2}{{10}}}}left( {{{100}^{frac{{ – 7}}{{10}}}}} right) hfill \
hfill \
= {100^{frac{2}{{10}} + left( { – frac{7}{{10}}} right)}} hfill \
= {100^{ – frac{1}{2}}} = frac{1}{{{{100}^{frac{1}{2}}}}} hfill \
= frac{1}{{sqrt {100} }} = frac{1}{{10}} hfill \
hfill \
end{gathered} ]
{text{c) Remember that }}frac{{{a^m}}}{{{a^n}}} = {a^{m – n}}{text{ }} hfill \
hfill \
{text{ }}frac{{{{64}^{frac{4}{3}}}}}{{64}} = {64^{frac{4}{3} – 1}} = {64^{frac{1}{3}}} hfill \
= sqrt[3]{{64}} = 4 hfill \
end{gathered} ]
{text{d) Remember that }}frac{{{a^m}}}{{{a^n}}} = {a^{m – n}}{text{ }} hfill \
{text{ }}frac{{{{27}^{ – 1}}}}{{{{27}^{ – frac{2}{3}}}}} = {27^{ – 1 – left( {frac{2}{3}} right)}} hfill \
= {27^{ – frac{1}{3}}} = frac{1}{{{{27}^{frac{1}{3}}}}} hfill \
= frac{1}{{sqrt[3]{{27}}}} = frac{1}{{sqrt[3]{{{3^3}}}}} = frac{1}{3} hfill \
end{gathered} ]
{text{e) Remember the following rules}} hfill \
{text{ }}{left( {{a^m}} right)^n} = {a^{m times n}}{text{ }} hfill \
{text{ }}frac{{{a^m}}}{{{a^n}}} = {a^{m – n}} hfill \
hfill \
frac{{{{left( {{text{1}}{{text{6}}^{ – 2.5}}} right)}^{ – 0.2}}}}{{{{16}^{frac{3}{4}}}}} = frac{{{{16}^{ – 2.5left( { – 0.2} right)}}}}{{{{16}^{frac{3}{4}}}}} hfill \
= frac{{{{16}^{ – 0.5}}}}{{{{16}^{frac{3}{4}}}}} = {16^{ – frac{1}{2} – }}^{frac{3}{4}} hfill \
= {16^{ – frac{1}{4}}} = frac{1}{{{{16}^{frac{1}{4}}}}} hfill \
= frac{1}{{sqrt[4]{{16}}}} = frac{1}{{sqrt[4]{{{2^4}}}}} = frac{1}{2} hfill \
end{gathered} ]
{text{f) Remember the following rules}} hfill \
{a^m} cdot {a^n} = {a^{m + n}} hfill \
{text{ }}{left( {{a^m}} right)^n} = {a^{m times n}}{text{ }} hfill \
{text{ }}frac{{{a^m}}}{{{a^n}}} = {a^{m – n}} hfill \
hfill \
frac{{left( {{8^{ – 2}}} right)left( {{8^{2.5}}} right)}}{{{{left( {{8^{ – 6}}} right)}^{ – 0.25}}}} = frac{{{8^{ – 2 + 2.5}}}}{{{8^{ – 6left( { – 0.25} right)}}}} hfill \
= frac{{{8^{0.5}}}}{{{8^{ – 1.5}}}} = {8^{0.5 – left( { – 1.5} right)}} hfill \
= {8^2} = 64 hfill \
end{gathered} ]
b) $100^{-frac{1}{2}}=dfrac{1}{10}$
c) $64^{frac{1}{3}}=4$
d) $27^{-frac{1}{3}}=dfrac{1}{3}$
e) $16^{-frac{1}{4}}=dfrac{1}{2}$
f) $8^2=64$
{text{The prediction of order: c,d,a,e,b,f}} hfill \
{text{Remember this rule, }}{a^{frac{n}{m}}} = sqrt[m]{{{x^n}}} = {left( {sqrt[m]{x}} right)^n} hfill \
{text{a)}}sqrt[4]{{623}} = 4.996 hfill \
{text{b) 12}}{{text{5}}^{frac{2}{5}}} = 6.899 hfill \
{text{c) }}sqrt[{10}]{{10.24}} = 1.262 hfill \
{text{d) 80}}{text{.}}{{text{9}}^{frac{1}{4}}} = 2.999 hfill \
{text{e) 17}}{text{.}}{{text{5}}^{frac{5}{8}}} = 5.983 hfill \
{text{f) 21}}{text{.}}{{text{4}}^{frac{3}{2}}} = 98.997 hfill \
end{gathered} ]
b) $6.899$
c) $1.262$
d) $2.999$
e) $5.983$
f) $98.997$
{text{The volume of a cube }}V{text{ with side }}s{text{ is }}V = {s^3}.{text{ }} hfill \
{text{The length of each side is }}s = sqrt[3]{V}{text{ }} hfill \
s = sqrt[3]{V} hfill \
s = sqrt[3]{{0.015625}} hfill \
s = 0.25 hfill \
{text{Therefore 0}}{text{.25 m is the length of each side of a cube}}{text{.}} hfill \
end{gathered} ]
$27^{4/3}=81$
$27^{1.333}=80.991 101 73$
The values are close but NOT equal since $dfrac{4}{3}neq 1.333$
$27^{1.333}=80.991 101 73$
Close but not equal since $dfrac{4}{3}ne 1.333$
$(-100)^{0.2}=(-100)^{1/5}=sqrt[5]{-100}implies$ defined since 5 is odd
$(-100)^{0.5}=(-100)^{1/2}=sqrt[2]{-100}implies$ undefined since 2 is even
$(-100)^{0.5}=(-100)^{1/2}=sqrt[2]{-100}implies$ undefined since 2 is even
125^{-2/3}=dfrac{1}{125^{2/3}}
$$
$$
a^{-n}=dfrac{1}{a^n}
$$
dfrac{1}{125^{2/3}}=dfrac{1}{(sqrt[3]{125})^2}
$$
$$
a^{n/m}=(sqrt[m]{a})^n
$$
dfrac{1}{(sqrt[3]{125})^2}=dfrac{1}{sqrt[3]{(5^3)^2}}
$$
$$
125=5^3
$$
dfrac{1}{(sqrt[3]{5^3})^2}=dfrac{1}{5^2}=dfrac{1}{25}
$$
$$
sqrt[m]{a^m}=a
$$
125^{-2/3}=dfrac{1}{125^{2/3}}=dfrac{1}{(sqrt[3]{125})^2}=dfrac{1}{(sqrt[3]{5^3})^2}=dfrac{1}{5^2}=dfrac{1}{25}
$$
a) Observe that $0.375=dfrac{8}{3}$ and $256=2^{8}$
$-256^{0.375}=-(2^8)^{frac{3}{8}}$
Use the rule $(a^m)^n=a^{mtimes n}$
$$
=-(2^{8times frac{3}{8}})=-2^3=-8
$$
$15.625=15+0.625=15+dfrac{5}{8}=dfrac{15(8)+5}{8}=dfrac{125}{8}$
$125=5^3$ and $8=2^3$
$15.625^{frac{4}{3}}=left(dfrac{125}{8}right)^{frac{4}{3}}$
$=left(dfrac{5^3}{2^3}right)^{frac{4}{3}}$
$$
=left[left(dfrac{5}{2}right)^3right]^{frac{4}{3}}=left(dfrac{5}{2}right)^{4}=dfrac{5^4}{2^4}=dfrac{625}{16}=39.0625
$$
$-0.027=-dfrac{27}{1000}=-dfrac{3^3}{10^3}$
$sqrt[3]{-0.027^4}=sqrt[3]{-left(dfrac{3^3}{10^3}right)^4}$
Use the rule: $sqrt[m]{a^n}=a^{frac{n}{m}}$
$$
=-left[left(dfrac{3}{10}right)^3right]^{4/3}
$$
$$
=-left(dfrac{3}{10}right)^4=-dfrac{3^4}{10^4}=dfrac{81}{10;000}=0.0081
$$
$3.375=3+0.375=left(3+dfrac{3}{8}right)=dfrac{3(8)+3}{8}=dfrac{27}{8}$
$$
=(-3.375)^{frac{2}{3}}=left(-dfrac{27}{8}right)^{frac{2}{3}}=left(-dfrac{3^3}{2^3}right)^{2/3}=left(-dfrac{3}{2}right)^2=dfrac{9}{4}=2.25
$$
$0.0016=dfrac{16}{10;000}=dfrac{16}{625(16)}=dfrac{1}{625}=dfrac{1}{5^4}$
Use the rule $sqrt[m]{a^n}=a^{frac{n}{m}}$
$$
sqrt[4]{(0.0016)^3}=0.0016^{frac{3}{4}}=left(dfrac{1}{625}right)^{3/4}=left(dfrac{1}{5^4}right)^{3/4}=dfrac{1}{5^3}=dfrac{1}{125}=0.008
$$
$1.6=1+dfrac{6}{10}=dfrac{10+6}{10}=dfrac{16}{10}=dfrac{8}{5}$
$7776=6^5$
$$
(-7776)^{1.6}=(-6^5)^{8/5}=(-6)^8=1;679;616
$$
b) $39.0625$
c) $0.0081$
d) $2.25$
e) $0.008$
f) $1;679;616$
Then use the rule $a^{frac{n}{m}}=sqrt[m]{a^n}$
$4^{2.5}=4^{frac{5}{2}}=(4^5)^{frac{1}{2}}=sqrt{4^5}$
Thus, $4^{2.5}$ means the square root of 4 multiplied by itself 5 times.
a) $9^{frac{1}{2}}+4^{frac{1}{2}}=sqrt{9}+sqrt{4}=3+2=5$
$(9+4)^{frac{1}{2}}=13^{frac{1}{2}}=sqrt{13}$
Thus, $9^{frac{1}{2}}+4^{frac{1}{2}}neq(9+4)^{frac{1}{2}}$
The statement is FALSE.
$(9times4)^{frac{1}{2}}=36^{frac{1}{2}}=sqrt{36}=6$
Thus, $9^{frac{1}{2}}+4^{frac{1}{2}}ne(9times4)^{frac{1}{2}}$
The statement is FALSE.
Thus, the statement is FALSE.
Thus, the statement is TRUE.
Furthermore, we can also show that
$left(x^{frac{1}{3}}+y^{frac{1}{3}}right)^6=left(x^{frac{1}{3}}+y^{frac{1}{3}}right)^3left(x^{frac{1}{3}}+y^{frac{1}{3}}right)^3$
Remember the special product $(a+b)^3=a^3+3a^2b+3ab^2+b^3$
Since all terms are positive, the final answer should contain more than two terms. Thus,
$left(x^{frac{1}{3}}+y^{frac{1}{3}}right)^6ne x^2+y^2$
The statement is FALSE.
$left[left(x^{frac{1}{3}}right)left(y^{frac{1}{3}}right)right]^6=x^{frac{6}{3}}y^{frac{6}{3}}=x^2y^2$
Thus, the statement is TRUE.
b) false
c) false
d) true
e) false
f) true
If $a>0$, $a^{frac{m}{n}}$ is undefined when $n=0$
If $a<0$, $a^{frac{m}{n}}$ is undefined when $n=0$ or when $n$ is even number.
a) In this case, we have a negative base $(a<0)$, thus it is undefined when $n=0$ or when $n$ is even number and $mneq 0$. If $m=0$, then it will be undefined only when $n=0$.
b) Since $6$ is positive, it is undefined only when $n=0$.
b) undefined when $n=0$
By inspection, notice that $2^4=16$ and $4^2=16$
Thus, solutions could be $x=2$ and $y=4$ or $y=4$ and $x=2$
By inspection, we also see that the equation is always true when $x=y$ except for values of $x$ that makes $x^x$ undefined such as $x=-dfrac{1}{n}$ where $n$ is even number.
Let $y=tx$
$x^{tx}=(tx)^x$
$(x^t)^x=(tx)^x$
Raise both side to power $frac{1}{x}$
$x^t=tx$
Solve for $x$ in terms of $t$
$dfrac{x^t}{x}=t$
$x^{t-1}=t$
$x=t^{frac{1}{t-1}}$
Since $y=tx$
$y=tcdot t^{frac{1}{t-1}}$
$y=t^{frac{t}{t-1}}$
$left( t^{frac{1}{t-1}},t^{frac{t}{t-1}}right)$ for any positive $tneq 1$.
For example, when $t=2$
$x=2^{frac{1}{2-1}}=2$
$y=2^{frac{2}{2-1}}=4$
when $t=3$
$x=3^{frac{1}{3-1}}=3^{frac{1}{2}}=sqrt{3}$
$y=tx=3times sqrt{3}=3sqrt{3}$
Notice from the graph that $x=y$ is also a solution as we pointed out earlier.
$1.225=(1+i)^{12}$
Take the 12th root of both sides
$1.225^{frac{1}{12}}=left[(1+i)^{12}right]^{frac{1}{12}}$
$1.225^{frac{1}{12}}=1+i$
$$
i=1.225^{frac{1}{12}}-1=0.017
$$
$left(dfrac{1}{16}right)^{frac{1}{4}}-sqrt[3]{dfrac{8}{27}}=sqrt{x^2}$
$left(dfrac{1}{2^4}right)^{frac{1}{4}}-sqrt[3]{dfrac{2^3}{3^3}}= |x|$
$dfrac{1}{2}-dfrac{2}{3}= |x|$
$dfrac{3}{6}-dfrac{4}{6}= |x|$
$-dfrac{1}{6}= |x|$
No solution since $|x|$ cannot be negative.
$sqrt[3]{dfrac{1}{8}}-sqrt[4]{x^4}+15=sqrt[4]{16}$
$sqrt[3]{dfrac{1}{2^3}}-|x|+15=sqrt[4]{2^4}$
$dfrac{1}{2}-|x|+15=2$
$|x|=dfrac{1}{2}+15-2$
$|x|=dfrac{27}{2}$
The principal value of $x$ is
$$
x=dfrac{27}{2}
$$
b) $x=dfrac{27}{2}$
