Question

$$\frac { 32 ^ { 0.2 } + 81 ^ { 0.25 } = 1 } { 256 ^ { 0.5 } - 121 ^ { 0.5 } }$$

Answer

$$t=1/(e^2*h*v*a*l*u)$$

Solution


Rewrite \(32\) as \({2}^{5}\).
\[thevalue\times \frac{{({2}^{5})}^{0.2}+{81}^{0.25}}{{256}^{0.5}-{121}^{0.5}}=1\]
Use this rule: \({({x}^{a})}^{b}={x}^{ab}\).
\[thevalue\times \frac{2+{81}^{0.25}}{{256}^{0.5}-{121}^{0.5}}=1\]
Rewrite \(81\) as \({3}^{4}\).
\[thevalue\times \frac{2+{({3}^{4})}^{0.25}}{{256}^{0.5}-{121}^{0.5}}=1\]
Use this rule: \({({x}^{a})}^{b}={x}^{ab}\).
\[thevalue\times \frac{2+3}{{256}^{0.5}-{121}^{0.5}}=1\]
Simplify  \(2+3\)  to  \(5\).
\[thevalue\times \frac{5}{{256}^{0.5}-{121}^{0.5}}=1\]
Rewrite \(256\) as \({2}^{8}\).
\[thevalue\times \frac{5}{{({2}^{8})}^{0.5}-{121}^{0.5}}=1\]
Use this rule: \({({x}^{a})}^{b}={x}^{ab}\).
\[thevalue\times \frac{5}{{2}^{4}-{121}^{0.5}}=1\]
Simplify  \({2}^{4}\)  to  \(16\).
\[thevalue\times \frac{5}{16-{121}^{0.5}}=1\]
Rewrite \(121\) as \({11}^{2}\).
\[thevalue\times \frac{5}{16-{({11}^{2})}^{0.5}}=1\]
Use this rule: \({({x}^{a})}^{b}={x}^{ab}\).
\[thevalue\times \frac{5}{16-11}=1\]
Simplify  \(16-11\)  to  \(5\).
\[thevalue\times \frac{5}{5}=1\]
Cancel \(5\).
\[thevalue=1\]
Use Product Rule: \({x}^{a}{x}^{b}={x}^{a+b}\).
\[th{e}^{2}valu=1\]
Regroup terms.
\[{e}^{2}thvalu=1\]
Divide both sides by \({e}^{2}\).
\[thvalu=\frac{1}{{e}^{2}}\]
Divide both sides by \(h\).
\[tvalu=\frac{\frac{1}{{e}^{2}}}{h}\]
Simplify  \(\frac{\frac{1}{{e}^{2}}}{h}\)  to  \(\frac{1}{{e}^{2}h}\).
\[tvalu=\frac{1}{{e}^{2}h}\]
Divide both sides by \(v\).
\[talu=\frac{\frac{1}{{e}^{2}h}}{v}\]
Simplify  \(\frac{\frac{1}{{e}^{2}h}}{v}\)  to  \(\frac{1}{{e}^{2}hv}\).
\[talu=\frac{1}{{e}^{2}hv}\]
Divide both sides by \(a\).
\[tlu=\frac{\frac{1}{{e}^{2}hv}}{a}\]
Simplify  \(\frac{\frac{1}{{e}^{2}hv}}{a}\)  to  \(\frac{1}{{e}^{2}hva}\).
\[tlu=\frac{1}{{e}^{2}hva}\]
Divide both sides by \(l\).
\[tu=\frac{\frac{1}{{e}^{2}hva}}{l}\]
Simplify  \(\frac{\frac{1}{{e}^{2}hva}}{l}\)  to  \(\frac{1}{{e}^{2}hval}\).
\[tu=\frac{1}{{e}^{2}hval}\]
Divide both sides by \(u\).
\[t=\frac{\frac{1}{{e}^{2}hval}}{u}\]
Simplify  \(\frac{\frac{1}{{e}^{2}hval}}{u}\)  to  \(\frac{1}{{e}^{2}hvalu}\).
\[t=\frac{1}{{e}^{2}hvalu}\]