Question

Find the roots of the function f(x) = 3^x · (log2 (x) − 3)5 · e ^(x ^2−3x ).

Answer 1

Answer:

l do not know.......... sorry frnd......

Answer 2

Answer:

The root of the provided function is x=8

Step-by-step explanation:

Consider the provided function.

$f(x) = 3^x \cdot (log_2 x-3)5 \cdot e ^{(x ^2-3x )}$

We need to find the roots of the function.

To find the roots of the function substitute f(x)=0 in above function.

$3^x \cdot (log_2 x-3)5 \cdot e ^{(x ^2-3x )}=0$

Now apply zero product rule:

$\text{If}\:ab=0\:\mathrm{then}\:a=0\:\mathrm{or}\:b=0\:\left(\mathrm{or\:both}\:a=0\:\mathrm{and}\:b=0\right)$

Therefore,

$3^x=0$, $(log_2 x-3)=0$ or $5 \cdot e ^{(x ^2-3x )}=0$

As we know $a^{f\left(x\right)}\mathrm{\:cannot\:be\:zero\:or\:negative\:for\:}x\in \mathbb{R}$

Thus, $3^x=0$ and $5 \cdot e ^{(x ^2-3x )}=0$ has no solution.

Now solve for $(log2 (x)-3)=0$

$log_2 x=3$

Now use the property: $\mathrm{If}\:\log_a\left(b\right)=c\:\mathrm{then}\:b=a^c$

$x=2^3$

$x=8$

Hence, the root of the provided function is x=8