Question
Solve the inequality
Solve the inequality by testing the values in the interval
Solve the inequality by separating into cases
x∈(−∞,0)∪(23,+∞)
Evaluate
x5(2x−3)5(−x×7)3(3x×8)<0
Remove the parentheses
x5(2x−3)5(−x×7)3×3x×8<0
Simplify
More Steps

Evaluate
x5(2x−3)5(−x×7)3×3x×8
Use the commutative property to reorder the terms
x5(2x−3)5(−7x)3×3x×8
Multiply the terms with the same base by adding their exponents
x5+1(2x−3)5(−7x)3×3×8
Add the numbers
x6(2x−3)5(−7x)3×3×8
Multiply the terms
x6(2x−3)5(−7x)3×24
Simplify
−343x6(2x−3)5x3×24
Multiply the terms
−8232x6(2x−3)5x3
−8232x6(2x−3)5x3<0
Change the signs on both sides of the inequality and flip the inequality sign
8232x6(2x−3)5x3>0
Rewrite the expression
8232x6(2x−3)5x3=0
Elimination the left coefficient
x6(2x−3)5x3=0
Separate the equation into 3 possible cases
x6=0(2x−3)5=0x3=0
The only way a power can be 0 is when the base equals 0
x=0(2x−3)5=0x3=0
Solve the equation
More Steps

Evaluate
(2x−3)5=0
The only way a power can be 0 is when the base equals 0
2x−3=0
Move the constant to the right-hand side and change its sign
2x=0+3
Removing 0 doesn't change the value,so remove it from the expression
2x=3
Divide both sides
22x=23
Divide the numbers
x=23
x=0x=23x3=0
The only way a power can be 0 is when the base equals 0
x=0x=23x=0
Find the union
x=0x=23
Determine the test intervals using the critical values
x<00<x<23x>23
Choose a value form each interval
x1=−1x2=1x3=3
To determine if x<0 is the solution to the inequality,test if the chosen value x=−1 satisfies the initial inequality
More Steps

Evaluate
8232(−1)6(2(−1)−3)5(−1)3>0
Simplify
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Evaluate
8232(−1)6(2(−1)−3)5(−1)3
Evaluate the power
8232×1×(2(−1)−3)5(−1)3
Simplify
8232×1×(−2−3)5(−1)3
Subtract the numbers
8232×1×(−5)5(−1)3
Rewrite the expression
8232(−5)5(−1)3
Multiply the terms
−25725000(−1)3
Evaluate the power
−25725000(−1)
Multiply the numbers
25725000
25725000>0
Calculate
2.5725×107>0
Check the inequality
true
x<0 is the solutionx2=1x3=3
To determine if 0<x<23 is the solution to the inequality,test if the chosen value x=1 satisfies the initial inequality
More Steps

Evaluate
8232×16×(2×1−3)5×13>0
Simplify
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Evaluate
8232×16×(2×1−3)5×13
1 raised to any power equals to 1
8232×1×(2×1−3)5×13
Any expression multiplied by 1 remains the same
8232×1×(2−3)5×13
Subtract the numbers
8232×1×(−1)5×13
1 raised to any power equals to 1
8232×1×(−1)5×1
Rewrite the expression
8232(−1)5
Evaluate the power
8232(−1)
Multiply the numbers
−8232
−8232>0
Check the inequality
false
x<0 is the solution0<x<23 is not a solutionx3=3
To determine if x>23 is the solution to the inequality,test if the chosen value x=3 satisfies the initial inequality
More Steps

Evaluate
8232×36(2×3−3)5×33>0
Simplify
More Steps

Evaluate
8232×36(2×3−3)5×33
Multiply the numbers
8232×36(6−3)5×33
Subtract the numbers
8232×36×35×33
Multiply the terms with the same base by adding their exponents
8232×36+5+3
Add the numbers
8232×314
8232×314>0
Calculate
3.93734×1010>0
Check the inequality
true
x<0 is the solution0<x<23 is not a solutionx>23 is the solution
Solution
x∈(−∞,0)∪(23,+∞)
Show Solution
