Question 88626
I agree with the solution by ankor (Carl).  Here's another way of looking at it:
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Given:
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{{{ln(5-x) = 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(a,N)= 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^y = 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^12 = 5 – x}}}
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Since e = 2.718281828 this equation becomes:
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{{{(2.718281828)^12 = 5 – 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 – x}}}
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Adding x to both sides converts this equation to:
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{{{x + 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(5 +162749.7914) = 12}}}
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{{{ln(162754.7914) = 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.
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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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