Basic Math Examples

Solve for k 1534/1142=e^(2k)
15341142=e2k15341142=e2k
Step 1
Rewrite the equation as e2k=15341142e2k=15341142.
e2k=15341142e2k=15341142
Step 2
Take the natural logarithm of both sides of the equation to remove the variable from the exponent.
ln(e2k)=ln(15341142)ln(e2k)=ln(15341142)
Step 3
Expand the left side.
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Step 3.1
Expand ln(e2k)ln(e2k) by moving 2k2k outside the logarithm.
2kln(e)=ln(15341142)2kln(e)=ln(15341142)
Step 3.2
The natural logarithm of ee is 11.
2k1=ln(15341142)2k1=ln(15341142)
Step 3.3
Multiply 22 by 11.
2k=ln(15341142)2k=ln(15341142)
2k=ln(15341142)2k=ln(15341142)
Step 4
Simplify the right side.
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Step 4.1
Cancel the common factor of 15341534 and 11421142.
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Step 4.1.1
Factor 22 out of 15341534.
2k=ln(2(767)1142)2k=ln(2(767)1142)
Step 4.1.2
Cancel the common factors.
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Step 4.1.2.1
Factor 22 out of 11421142.
2k=ln(27672571)2k=ln(27672571)
Step 4.1.2.2
Cancel the common factor.
2k=ln(27672571)
Step 4.1.2.3
Rewrite the expression.
2k=ln(767571)
2k=ln(767571)
2k=ln(767571)
2k=ln(767571)
Step 5
Divide each term in 2k=ln(767571) by 2 and simplify.
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Step 5.1
Divide each term in 2k=ln(767571) by 2.
2k2=ln(767571)2
Step 5.2
Simplify the left side.
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Step 5.2.1
Cancel the common factor of 2.
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Step 5.2.1.1
Cancel the common factor.
2k2=ln(767571)2
Step 5.2.1.2
Divide k by 1.
k=ln(767571)2
k=ln(767571)2
k=ln(767571)2
k=ln(767571)2
Step 6
The result can be shown in multiple forms.
Exact Form:
k=ln(767571)2
Decimal Form:
k=0.14754879
 [x2  12  π  xdx ]