$(x^2-3x+1)^{(2x^2+x-6)}=1\ we \ take \ the \ logarithm\\\\(2x^2+x-6)*ln(x^2-3x+1)=0\\\\(2x^2+x-6)=0\ or\ x^2-3x+1=1\\\\1)\\2x^2+x-6=0\\2x^2+4x-3x-6=0\\2x(x+2)-3(x+2)=0\\(x+2)(2x-3)\\x=-2\ or\ x=\dfrac{3}{2} \\\\2)\\x^2-3x=0\\x(x-3)=0\\x=0\ or\ x=3\\\\sol=\{-2,0,\dfrac{3}{2},3\}\\$

8 0

3 years ago

Help help help help help

shtirl [24]

Answer:

Step-by-step explanation:

6 0

2 years ago

Read 2 more answers

Please find the domain and range!

SpyIntel [72]

Answer:

Domain: {-2, -3, 6, 8, 10}

Range: {-5, 1, 7, 9}

Step-by-step explanation:

Given:

{(6, -5), (-2, 9), (-3, 1), (10, 7), (8, 9)}

✔️Domain:

This includes all the set of the x-values that are in the relation. This includes, 6, -2, -3, 10, and 8.

Thus, the domain can be represented as:

{-2, -3, 6, 8, 10}

✔️Range:

This includes all corresponding y-values in the relation. They are, -5, 1, 7, and 9.

Range can be represented as:

{-5, 1, 7, 9}

3 0

3 years ago