Finite Math Examples

Solve for x natural log of natural log of x-e^6x=0
ln(ln(x-e6x))=0ln(ln(xe6x))=0
Step 1
To solve for x, rewrite the equation using properties of logarithms.
eln(ln(x-e6x))=e0
Step 2
Rewrite ln(ln(x-e6x))=0 in exponential form using the definition of a logarithm. If x and b are positive real numbers and b1, then logb(x)=y is equivalent to by=x.
e0=ln(x-e6x)
Step 3
Solve for x.
Tap for more steps...
Step 3.1
Rewrite the equation as ln(x-e6x)=e0.
ln(x-e6x)=e0
Step 3.2
To solve for x, rewrite the equation using properties of logarithms.
eln(x-e6x)=ee0
Step 3.3
Rewrite ln(x-e6x)=e0 in exponential form using the definition of a logarithm. If x and b are positive real numbers and b1, then logb(x)=y is equivalent to by=x.
ee0=x-e6x
Step 3.4
Solve for x.
Tap for more steps...
Step 3.4.1
Rewrite the equation as x-e6x=ee0.
x-e6x=ee0
Step 3.4.2
Simplify ee0.
Tap for more steps...
Step 3.4.2.1
Anything raised to 0 is 1.
x-e6x=e1
Step 3.4.2.2
Simplify.
x-e6x=e
x-e6x=e
Step 3.4.3
Factor the left side of the equation.
Tap for more steps...
Step 3.4.3.1
Factor x out of x-e6x.
Tap for more steps...
Step 3.4.3.1.1
Raise x to the power of 1.
x-e6x=e
Step 3.4.3.1.2
Factor x out of x1.
x1-e6x=e
Step 3.4.3.1.3
Factor x out of -e6x.
x1+x(-e6)=e
Step 3.4.3.1.4
Factor x out of x1+x(-e6).
x(1-e6)=e
x(1-e6)=e
Step 3.4.3.2
Rewrite 1 as 13.
x(13-e6)=e
Step 3.4.3.3
Rewrite e6 as (e2)3.
x(13-(e2)3)=e
Step 3.4.3.4
Since both terms are perfect cubes, factor using the difference of cubes formula, a3-b3=(a-b)(a2+ab+b2) where a=1 and b=e2.
x((1-e2)(12+1e2+(e2)2))=e
Step 3.4.3.5
Factor.
Tap for more steps...
Step 3.4.3.5.1
Simplify.
Tap for more steps...
Step 3.4.3.5.1.1
Rewrite 1 as 12.
x((12-e2)(12+1e2+(e2)2))=e
Step 3.4.3.5.1.2
Since both terms are perfect squares, factor using the difference of squares formula, a2-b2=(a+b)(a-b) where a=1 and b=e.
x((1+e)(1-e)(12+1e2+(e2)2))=e
Step 3.4.3.5.1.3
Multiply e2 by 1.
x((1+e)(1-e)(12+e2+(e2)2))=e
x((1+e)(1-e)(12+e2+(e2)2))=e
Step 3.4.3.5.2
Remove unnecessary parentheses.
x(1+e)(1-e)(12+e2+(e2)2)=e
x(1+e)(1-e)(12+e2+(e2)2)=e
Step 3.4.3.6
One to any power is one.
x(1+e)(1-e)(1+e2+(e2)2)=e
Step 3.4.3.7
Multiply the exponents in (e2)2.
Tap for more steps...
Step 3.4.3.7.1
Apply the power rule and multiply exponents, (am)n=amn.
x(1+e)(1-e)(1+e2+e22)=e
Step 3.4.3.7.2
Multiply 2 by 2.
x(1+e)(1-e)(1+e2+e4)=e
x(1+e)(1-e)(1+e2+e4)=e
x(1+e)(1-e)(1+e2+e4)=e
Step 3.4.4
Divide each term in x(1+e)(1-e)(1+e2+e4)=e by 1-e6 and simplify.
Tap for more steps...
Step 3.4.4.1
Divide each term in x(1+e)(1-e)(1+e2+e4)=e by 1-e6.
x(1+e)(1-e)(1+e2+e4)1-e6=e1-e6
Step 3.4.4.2
Simplify the left side.
Tap for more steps...
Step 3.4.4.2.1
Simplify the denominator.
Tap for more steps...
Step 3.4.4.2.1.1
Rewrite 1 as 13.
x(1+e)(1-e)(1+e2+e4)13-e6=e1-e6
Step 3.4.4.2.1.2
Rewrite e6 as (e2)3.
x(1+e)(1-e)(1+e2+e4)13-(e2)3=e1-e6
Step 3.4.4.2.1.3
Since both terms are perfect cubes, factor using the difference of cubes formula, a3-b3=(a-b)(a2+ab+b2) where a=1 and b=e2.
x(1+e)(1-e)(1+e2+e4)(1-e2)(12+1e2+(e2)2)=e1-e6
Step 3.4.4.2.1.4
Simplify.
Tap for more steps...
Step 3.4.4.2.1.4.1
Rewrite 1 as 12.
x(1+e)(1-e)(1+e2+e4)(12-e2)(12+1e2+(e2)2)=e1-e6
Step 3.4.4.2.1.4.2
Since both terms are perfect squares, factor using the difference of squares formula, a2-b2=(a+b)(a-b) where a=1 and b=e.
x(1+e)(1-e)(1+e2+e4)(1+e)(1-e)(12+1e2+(e2)2)=e1-e6
Step 3.4.4.2.1.4.3
Multiply e2 by 1.
x(1+e)(1-e)(1+e2+e4)(1+e)(1-e)(12+e2+(e2)2)=e1-e6
x(1+e)(1-e)(1+e2+e4)(1+e)(1-e)(12+e2+(e2)2)=e1-e6
Step 3.4.4.2.1.5
Simplify each term.
Tap for more steps...
Step 3.4.4.2.1.5.1
One to any power is one.
x(1+e)(1-e)(1+e2+e4)(1+e)(1-e)(1+e2+(e2)2)=e1-e6
Step 3.4.4.2.1.5.2
Multiply the exponents in (e2)2.
Tap for more steps...
Step 3.4.4.2.1.5.2.1
Apply the power rule and multiply exponents, (am)n=amn.
x(1+e)(1-e)(1+e2+e4)(1+e)(1-e)(1+e2+e22)=e1-e6
Step 3.4.4.2.1.5.2.2
Multiply 2 by 2.
x(1+e)(1-e)(1+e2+e4)(1+e)(1-e)(1+e2+e4)=e1-e6
x(1+e)(1-e)(1+e2+e4)(1+e)(1-e)(1+e2+e4)=e1-e6
x(1+e)(1-e)(1+e2+e4)(1+e)(1-e)(1+e2+e4)=e1-e6
x(1+e)(1-e)(1+e2+e4)(1+e)(1-e)(1+e2+e4)=e1-e6
Step 3.4.4.2.2
Reduce the expression by cancelling the common factors.
Tap for more steps...
Step 3.4.4.2.2.1
Cancel the common factor of 1+e.
Tap for more steps...
Step 3.4.4.2.2.1.1
Cancel the common factor.
x(1+e)(1-e)(1+e2+e4)(1+e)(1-e)(1+e2+e4)=e1-e6
Step 3.4.4.2.2.1.2
Rewrite the expression.
(x(1-e))(1+e2+e4)(1-e)(1+e2+e4)=e1-e6
(x(1-e))(1+e2+e4)(1-e)(1+e2+e4)=e1-e6
Step 3.4.4.2.2.2
Cancel the common factor of 1-e.
Tap for more steps...
Step 3.4.4.2.2.2.1
Cancel the common factor.
x(1-e)(1+e2+e4)(1-e)(1+e2+e4)=e1-e6
Step 3.4.4.2.2.2.2
Rewrite the expression.
(x)(1+e2+e4)1+e2+e4=e1-e6
(x)(1+e2+e4)1+e2+e4=e1-e6
Step 3.4.4.2.2.3
Cancel the common factor of 1+e2+e4.
Tap for more steps...
Step 3.4.4.2.2.3.1
Cancel the common factor.
x(1+e2+e4)1+e2+e4=e1-e6
Step 3.4.4.2.2.3.2
Divide x by 1.
x=e1-e6
x=e1-e6
x=e1-e6
x=e1-e6
Step 3.4.4.3
Simplify the right side.
Tap for more steps...
Step 3.4.4.3.1
Simplify the denominator.
Tap for more steps...
Step 3.4.4.3.1.1
Rewrite 1 as 13.
x=e13-e6
Step 3.4.4.3.1.2
Rewrite e6 as (e2)3.
x=e13-(e2)3
Step 3.4.4.3.1.3
Since both terms are perfect cubes, factor using the difference of cubes formula, a3-b3=(a-b)(a2+ab+b2) where a=1 and b=e2.
x=e(1-e2)(12+1e2+(e2)2)
Step 3.4.4.3.1.4
Simplify.
Tap for more steps...
Step 3.4.4.3.1.4.1
Rewrite 1 as 12.
x=e(12-e2)(12+1e2+(e2)2)
Step 3.4.4.3.1.4.2
Since both terms are perfect squares, factor using the difference of squares formula, a2-b2=(a+b)(a-b) where a=1 and b=e.
x=e(1+e)(1-e)(12+1e2+(e2)2)
Step 3.4.4.3.1.4.3
Multiply e2 by 1.
x=e(1+e)(1-e)(12+e2+(e2)2)
x=e(1+e)(1-e)(12+e2+(e2)2)
Step 3.4.4.3.1.5
Simplify each term.
Tap for more steps...
Step 3.4.4.3.1.5.1
One to any power is one.
x=e(1+e)(1-e)(1+e2+(e2)2)
Step 3.4.4.3.1.5.2
Multiply the exponents in (e2)2.
Tap for more steps...
Step 3.4.4.3.1.5.2.1
Apply the power rule and multiply exponents, (am)n=amn.
x=e(1+e)(1-e)(1+e2+e22)
Step 3.4.4.3.1.5.2.2
Multiply 2 by 2.
x=e(1+e)(1-e)(1+e2+e4)
x=e(1+e)(1-e)(1+e2+e4)
x=e(1+e)(1-e)(1+e2+e4)
x=e(1+e)(1-e)(1+e2+e4)
x=e(1+e)(1-e)(1+e2+e4)
x=e(1+e)(1-e)(1+e2+e4)
x=e(1+e)(1-e)(1+e2+e4)
x=e(1+e)(1-e)(1+e2+e4)
Step 4
The result can be shown in multiple forms.
Exact Form:
x=e(1+e)(1-e)(1+e2+e4)
Decimal Form:
x=-0.00675469
 [x2  12  π  xdx ]