How do you solve? log(x-1) + log 2x= log2x^2 - log2

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Nimo N. Share
Feb 19, 2018

Answer:

The solution is: # color(brown)( x = 2 #

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How do you solve
# color(blue)( log (x - 1) + log (2x) = log (2x^2) - log (2) #

  • Please note: It is assumed that an expression like log(x) means take the common logarithm, i.e. base 10 logarithm of x. If another base is intended, it should be indicated, as in # color(brown)( log_2 (x) #. or use # color(brown)( ln (x) # for the "natural, base e logarithm".

Using properties of logarithms, we can combine terms, then rearrange terms to isolate x, if possible. Also, we must assume # color(blue)( 1 < x #, otherwise the equation would not be valid for a real valued logarithm function.

# log (x - 1) + log (2x) = log (2x^2) - log (2) #
# log ( (x - 1) * (2x) ) = log ( (2x^2)/2 ) #

If two logarithm expressions with the same base are equal then the arguments of the functions are equal.
# (x - 1) * (2x) = ( (2x^2)/2 ) #
Simplify and solve for x.
# 2x^2 - 2x = x^2 #
Subtract # x^2 # from both sides.
# x^2 - 2x = 0 #
# x * (x - 2) = 0 #

Either # x = 0 #, which is not allowed,
or
# (x - 2) = 0 #
# color(brown)( x = 2 #

We're not finished until we check, since there may have been an error in the work, or the only possible solution may not work.
# log ((2) - 1) + log (2(2)) =^? log (2(2)^2) - log (2) #
# log (1) + log (4) =^? log (8) - log (2) #
# 0 + log (4) =^? log (8/2) #
# log (4) =^? log (4) #, YES!!! We have our solution.

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