Precalculus Examples

2x+3=32x+3=3
Step 1
Take the natural logarithm of both sides of the equation to remove the variable from the exponent.
ln(2x+3)=ln(3)ln(2x+3)=ln(3)
Step 2
Expand ln(2x+3)ln(2x+3) by moving x+3x+3 outside the logarithm.
(x+3)ln(2)=ln(3)(x+3)ln(2)=ln(3)
Step 3
Simplify the left side.
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Step 3.1
Apply the distributive property.
xln(2)+3ln(2)=ln(3)xln(2)+3ln(2)=ln(3)
xln(2)+3ln(2)=ln(3)xln(2)+3ln(2)=ln(3)
Step 4
Move all the terms containing a logarithm to the left side of the equation.
xln(2)+3ln(2)-ln(3)=0xln(2)+3ln(2)ln(3)=0
Step 5
Move all terms not containing xx to the right side of the equation.
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Step 5.1
Subtract 3ln(2)3ln(2) from both sides of the equation.
xln(2)-ln(3)=-3ln(2)xln(2)ln(3)=3ln(2)
Step 5.2
Add ln(3)ln(3) to both sides of the equation.
xln(2)=-3ln(2)+ln(3)xln(2)=3ln(2)+ln(3)
xln(2)=-3ln(2)+ln(3)xln(2)=3ln(2)+ln(3)
Step 6
Divide each term in xln(2)=-3ln(2)+ln(3)xln(2)=3ln(2)+ln(3) by ln(2)ln(2) and simplify.
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Step 6.1
Divide each term in xln(2)=-3ln(2)+ln(3)xln(2)=3ln(2)+ln(3) by ln(2)ln(2).
xln(2)ln(2)=-3ln(2)ln(2)+ln(3)ln(2)xln(2)ln(2)=3ln(2)ln(2)+ln(3)ln(2)
Step 6.2
Simplify the left side.
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Step 6.2.1
Cancel the common factor of ln(2)ln(2).
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Step 6.2.1.1
Cancel the common factor.
xln(2)ln(2)=-3ln(2)ln(2)+ln(3)ln(2)xln(2)ln(2)=3ln(2)ln(2)+ln(3)ln(2)
Step 6.2.1.2
Divide xx by 11.
x=-3ln(2)ln(2)+ln(3)ln(2)x=3ln(2)ln(2)+ln(3)ln(2)
x=-3ln(2)ln(2)+ln(3)ln(2)x=3ln(2)ln(2)+ln(3)ln(2)
x=-3ln(2)ln(2)+ln(3)ln(2)x=3ln(2)ln(2)+ln(3)ln(2)
Step 6.3
Simplify the right side.
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Step 6.3.1
Cancel the common factor of ln(2)ln(2).
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Step 6.3.1.1
Cancel the common factor.
x=-3ln(2)ln(2)+ln(3)ln(2)x=3ln(2)ln(2)+ln(3)ln(2)
Step 6.3.1.2
Divide -33 by 11.
x=-3+ln(3)ln(2)x=3+ln(3)ln(2)
x=-3+ln(3)ln(2)x=3+ln(3)ln(2)
x=-3+ln(3)ln(2)x=3+ln(3)ln(2)
x=-3+ln(3)ln(2)x=3+ln(3)ln(2)
Step 7
The result can be shown in multiple forms.
Exact Form:
x=-3+ln(3)ln(2)x=3+ln(3)ln(2)
Decimal Form:
x=-1.41503749x=1.41503749
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 [x2  12  π  xdx ]  x2  12  π  xdx  
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