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Finite Math Examples
log(7√x)-log(log7((x)5))log(7√x)−log(log7((x)5))
Step 1
Set the argument in log(7√x)log(7√x) less than or equal to 00 to find where the expression is undefined.
7√x≤07√x≤0
Step 2
Step 2.1
To remove the radical on the left side of the inequality, raise both sides of the inequality to the power of 77.
7√x7≤077√x7≤07
Step 2.2
Simplify each side of the inequality.
Step 2.2.1
Use n√ax=axnn√ax=axn to rewrite 7√x7√x as x17x17.
(x17)7≤07(x17)7≤07
Step 2.2.2
Simplify the left side.
Step 2.2.2.1
Simplify (x17)7(x17)7.
Step 2.2.2.1.1
Multiply the exponents in (x17)7(x17)7.
Step 2.2.2.1.1.1
Apply the power rule and multiply exponents, (am)n=amn(am)n=amn.
x17⋅7≤07x17⋅7≤07
Step 2.2.2.1.1.2
Cancel the common factor of 77.
Step 2.2.2.1.1.2.1
Cancel the common factor.
x17⋅7≤07
Step 2.2.2.1.1.2.2
Rewrite the expression.
x1≤07
x1≤07
x1≤07
Step 2.2.2.1.2
Simplify.
x≤07
x≤07
x≤07
Step 2.2.3
Simplify the right side.
Step 2.2.3.1
Raising 0 to any positive power yields 0.
x≤0
x≤0
x≤0
x≤0
Step 3
Set the argument in log7((x)5) less than or equal to 0 to find where the expression is undefined.
(x)5≤0
Step 4
Step 4.1
Take the specified root of both sides of the inequality to eliminate the exponent on the left side.
5√x5≤5√0
Step 4.2
Simplify the equation.
Step 4.2.1
Simplify the left side.
Step 4.2.1.1
Pull terms out from under the radical.
x≤5√0
x≤5√0
Step 4.2.2
Simplify the right side.
Step 4.2.2.1
Simplify 5√0.
Step 4.2.2.1.1
Rewrite 0 as 05.
x≤5√05
Step 4.2.2.1.2
Pull terms out from under the radical.
x≤0
x≤0
x≤0
x≤0
x≤0
Step 5
Set the argument in log(log7((x)5)) less than or equal to 0 to find where the expression is undefined.
log7((x)5)≤0
Step 6
Step 6.1
Remove parentheses.
log7(x5)≤0
Step 6.2
Graph each side of the equation. The solution is the x-value of the point of intersection.
x=1
Step 6.3
Find the domain of log7((x)5).
Step 6.3.1
Set the argument in log7((x)5) greater than 0 to find where the expression is defined.
(x)5>0
Step 6.3.2
Solve for x.
Step 6.3.2.1
Take the specified root of both sides of the inequality to eliminate the exponent on the left side.
5√x5>5√0
Step 6.3.2.2
Simplify the equation.
Step 6.3.2.2.1
Simplify the left side.
Step 6.3.2.2.1.1
Pull terms out from under the radical.
x>5√0
x>5√0
Step 6.3.2.2.2
Simplify the right side.
Step 6.3.2.2.2.1
Simplify 5√0.
Step 6.3.2.2.2.1.1
Rewrite 0 as 05.
x>5√05
Step 6.3.2.2.2.1.2
Pull terms out from under the radical.
x>0
x>0
x>0
x>0
x>0
Step 6.3.3
The domain is all values of x that make the expression defined.
(0,∞)
(0,∞)
Step 6.4
Use each root to create test intervals.
x<0
0<x<1
x>1
Step 6.5
Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.
Step 6.5.1
Test a value on the interval x<0 to see if it makes the inequality true.
Step 6.5.1.1
Choose a value on the interval x<0 and see if this value makes the original inequality true.
x=-2
Step 6.5.1.2
Replace x with -2 in the original inequality.
log7((-2)5)≤0
Step 6.5.1.3
Determine if the inequality is true.
Step 6.5.1.3.1
The equation cannot be solved because it is undefined.
Undefined
Step 6.5.1.3.2
The left side has no solution, which means that the given statement is false.
False
False
False
Step 6.5.2
Test a value on the interval 0<x<1 to see if it makes the inequality true.
Step 6.5.2.1
Choose a value on the interval 0<x<1 and see if this value makes the original inequality true.
x=0.5
Step 6.5.2.2
Replace x with 0.5 in the original inequality.
log7((0.5)5)≤0
Step 6.5.2.3
The left side -1.78103593 is less than the right side 0, which means that the given statement is always true.
True
True
Step 6.5.3
Test a value on the interval x>1 to see if it makes the inequality true.
Step 6.5.3.1
Choose a value on the interval x>1 and see if this value makes the original inequality true.
x=4
Step 6.5.3.2
Replace x with 4 in the original inequality.
log7((4)5)≤0
Step 6.5.3.3
The left side 3.56207187 is greater than the right side 0, which means that the given statement is false.
False
False
Step 6.5.4
Compare the intervals to determine which ones satisfy the original inequality.
x<0 False
0<x<1 True
x>1 False
x<0 False
0<x<1 True
x>1 False
Step 6.6
The solution consists of all of the true intervals.
0<x≤1
0<x≤1
Step 7
The equation is undefined where the denominator equals 0, the argument of a square root is less than 0, or the argument of a logarithm is less than or equal to 0.
x≤1
(-∞,1]
Step 8