SOLUTION: Solve ln(5-x) = 12.

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Question 88626: Solve ln(5-x) = 12.
Found 2 solutions by ankor@dixie-net.com, bucky:
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
Solve ln(5-x) = 12.
:
Find the anti-log (e^) of both sides
5 - x = 162754.7914
-x = 162754.7914 - 5
-x = 162749.7914
x = -16749.7914
:
Check on a good calc: enter: ln(5-(-16749.7914)) = 12

Answer by bucky(2189) About Me  (Show Source):
You can put this solution on YOUR website!
I agree with the solution by ankor (Carl). Here's another way of looking at it:
.
Given:
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ln%285-x%29+=+12
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This equation can be solved by converting the logarithmic form to the exponential
form. This conversion is defined by the equation:
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log%28a%2CN%29=+y
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In which a is the base of the logarithm. By definition this logarithmic form is equivalent
to the exponential form:
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a%5Ey+=+N
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By comparing the logarithmic form to the given problem you can see that N = (5 – x) and
y = 12. The value of “a” is not quite as straightforward. Natural logarithms, designated by
the use of “ln” is equivalent to log to the base “e” in which e = 2.718181828. So
“a” = “e” = 2.718281828. If you substitute these given values into the exponential form
you get:
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e%5E12+=+5+%96+x
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Since e = 2.718281828 this equation becomes:
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%282.718281828%29%5E12+=+5+%96+x+
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A scientific calculator can be used to raise e to the 12th power. When you do that you get:
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162754.7914+=+5+%96+x
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Adding x to both sides converts this equation to:
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x+%2B+162754.7914+=+5+.
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Then subtracting 162754.7914 from both sides results in:
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+x+=+-162749.7914
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Check this answer by returning to the original equation and substituting -162749.7914
for x. This results in ln(5-(-162749.7914)) which becomes:
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ln%285+%2B162749.7914%29+=+12
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ln%28162754.7914%29+=+12
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And using a scientific calculator you can enter 162754.7914 and press the ln key to find
that ln(162754.7914) does in fact equal 12. So the answer for x is correct. It does
equal -162749.7914.
.
Hope this helps you to see another way ... transferring from logarithmic form to exponential
form is often a useful method in solving logarithmic equations.
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