Precalculus Examples

Solve for x e^x-30e^(-x)-1=0
ex-30e-x-1=0ex30ex1=0
Step 1
Rewrite e-xex as exponentiation.
ex-30(ex)-1-1=0ex30(ex)11=0
Step 2
Substitute uu for exex.
u-30u-1-1=0u30u11=0
Step 3
Simplify each term.
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Step 3.1
Rewrite the expression using the negative exponent rule b-n=1bnbn=1bn.
u-301u-1=0u301u1=0
Step 3.2
Combine -3030 and 1u1u.
u+-30u-1=0u+30u1=0
Step 3.3
Move the negative in front of the fraction.
u-30u-1=0u30u1=0
u-30u-1=0u30u1=0
Step 4
Solve for uu.
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Step 4.1
Find the LCD of the terms in the equation.
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Step 4.1.1
Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.
1,u,1,11,u,1,1
Step 4.1.2
The LCM of one and any expression is the expression.
uu
uu
Step 4.2
Multiply each term in u-30u-1=0u30u1=0 by uu to eliminate the fractions.
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Step 4.2.1
Multiply each term in u-30u-1=0u30u1=0 by uu.
uu-30uu-u=0uuu30uuu=0u
Step 4.2.2
Simplify the left side.
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Step 4.2.2.1
Simplify each term.
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Step 4.2.2.1.1
Multiply uu by uu.
u2-30uu-u=0uu230uuu=0u
Step 4.2.2.1.2
Cancel the common factor of uu.
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Step 4.2.2.1.2.1
Move the leading negative in -30u30u into the numerator.
u2+-30uu-u=0uu2+30uuu=0u
Step 4.2.2.1.2.2
Cancel the common factor.
u2+-30uu-u=0u
Step 4.2.2.1.2.3
Rewrite the expression.
u2-30-u=0u
u2-30-u=0u
u2-30-u=0u
u2-30-u=0u
Step 4.2.3
Simplify the right side.
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Step 4.2.3.1
Multiply 0 by u.
u2-30-u=0
u2-30-u=0
u2-30-u=0
Step 4.3
Solve the equation.
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Step 4.3.1
Factor u2-30-u using the AC method.
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Step 4.3.1.1
Consider the form x2+bx+c. Find a pair of integers whose product is c and whose sum is b. In this case, whose product is -30 and whose sum is -1.
-6,5
Step 4.3.1.2
Write the factored form using these integers.
(u-6)(u+5)=0
(u-6)(u+5)=0
Step 4.3.2
If any individual factor on the left side of the equation is equal to 0, the entire expression will be equal to 0.
u-6=0
u+5=0
Step 4.3.3
Set u-6 equal to 0 and solve for u.
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Step 4.3.3.1
Set u-6 equal to 0.
u-6=0
Step 4.3.3.2
Add 6 to both sides of the equation.
u=6
u=6
Step 4.3.4
Set u+5 equal to 0 and solve for u.
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Step 4.3.4.1
Set u+5 equal to 0.
u+5=0
Step 4.3.4.2
Subtract 5 from both sides of the equation.
u=-5
u=-5
Step 4.3.5
The final solution is all the values that make (u-6)(u+5)=0 true.
u=6,-5
u=6,-5
u=6,-5
Step 5
Substitute 6 for u in u=ex.
6=ex
Step 6
Solve 6=ex.
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Step 6.1
Rewrite the equation as ex=6.
ex=6
Step 6.2
Take the natural logarithm of both sides of the equation to remove the variable from the exponent.
ln(ex)=ln(6)
Step 6.3
Expand the left side.
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Step 6.3.1
Expand ln(ex) by moving x outside the logarithm.
xln(e)=ln(6)
Step 6.3.2
The natural logarithm of e is 1.
x1=ln(6)
Step 6.3.3
Multiply x by 1.
x=ln(6)
x=ln(6)
x=ln(6)
Step 7
Substitute -5 for u in u=ex.
-5=ex
Step 8
Solve -5=ex.
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Step 8.1
Rewrite the equation as ex=-5.
ex=-5
Step 8.2
Take the natural logarithm of both sides of the equation to remove the variable from the exponent.
ln(ex)=ln(-5)
Step 8.3
The equation cannot be solved because ln(-5) is undefined.
Undefined
Step 8.4
There is no solution for ex=-5
No solution
No solution
Step 9
List the solutions that makes the equation true.
x=ln(6)
Step 10
The result can be shown in multiple forms.
Exact Form:
x=ln(6)
Decimal Form:
x=1.79175946
 [x2  12  π  xdx ]