For this case, we must find the value of "x" so that the given expression is equal to 8.
That is to say:
![(\sqrt [5] {8 ^ 3}) ^ x = 8](https://img.qammunity.org/2020/formulas/mathematics/middle-school/dn2t62pqbtlx34pv4ax1kdqhcwcn25pazg.png)
We apply "ln" to both sides of the equation to remove the exponent variable:
![ln ((\sqrt [5] {8 ^ 3}) ^ x) = ln (8)\\xln (\sqrt [5] {8 ^ 3}) = ln (8)\\xln (\sqrt [5] {512}) = ln (8)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/zfz04u591aakcvg8up0oj9f4ok0k1ykao3.png)
We rewrite 512 as:
![512 = 32 * 16 = 2 ^ 5 * 16\\xln (\sqrt [5] {2 ^ 5 * 16}) = ln (8)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/nbxntji3od8b8d1uhn6b1ljv2ogsr8tae6.png)
By definition of power properties we have:
![\sqrt [n] {a ^ m} = a ^ {\frac {m} {n}}](https://img.qammunity.org/2020/formulas/mathematics/high-school/suo6ai2uezolc3t7f2f9e9h1lijquf271f.png)
So:
![xln (2 \sqrt [5] {16}) = ln (8)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/1o6lylpx6gzuj7ec4w3exijzrlk7shhux0.png)
We clear x:
![x = \frac {ln (8)} {ln (2 \sqrt [5] {16})}](https://img.qammunity.org/2020/formulas/mathematics/middle-school/o5f14nptvottaemm7h57po20qctja4i76m.png)
In decimal form,
periodic number
ANswer:
![x = \frac {ln (8)} {ln (2 \sqrt [5] {16})}\\x = 1.6\ periodic\ number](https://img.qammunity.org/2020/formulas/mathematics/middle-school/2koukgzq3t0vz0ocj78umkdxno69ciy510.png)