How do you solve #Log(3x–5)=3 #?

1 Answer
Alan P.
Dec 6, 2015

#x=331 2/3#

Explanation:

Two very important things to remember when working with #log# function

  1. #log(a)# means #log_(10) a#
    (The default base for the #log# function is #10#).

  2. #color(black)(log_b a = c)# means #color(black)(b^c=a)#
    (Of the two this is the one you really need to memorize and practice using).

Applying this to the given example:
#log(3x-5)=3#

means
#color(white)("XXX")log_10(3x-5)=3#

which in turn means
#color(white)("XXX")10^3 = 3x-5#

We can simplify this as
#color(white)("XXXXX")1000 = 3x-5#

#rarrcolor(white)("XXX")995 = 3x#

#rarrcolor(white)("XXX")x=331 2/3#

Again, let me emphasize:
the general equivalence
#color(white)("XXX")color(red)(log_b a =c <=> b^c=a)#
is something most people do not grasp intuitively and should be memorized and worked with until it comes naturally.

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