How do you solve #log_10(x+6) + log_10(3x-2)=2#?

1 Answer
Tony B
Dec 5, 2015

I have taken this to a point where you should be able to take over and complete the calculation. The final format is that of a quadratic.

Explanation:

The left hand side (LHS) is stating that you should take logs of 2 numbers. The RHS is just giving a value. So the RHS is the value you get after taking a log.

So before we start we need to ask the question; what is #x# if #log_10(x)=2#. The answer to that is 100.

Rewrite the given question as:

#log_10(x+6)+log_10(3x-2)=log_10(100).........(1)#

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

For logs to be added it means that the source is multiplied so we have:

Consider the LHS only

#log_10(x+6)+log_10(3x-2)-> log_10[color(white)(.)(x+5)(3x-2)color(white)(.)]#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Rewrite equation (1) as

#log_10[color(white)(.)(x+5)(3x-2)color(white)(.)]=log_10(200)#

#=> (x+5)(3x-2)=200#

Now you just multiply out the brackets and put it into standard form!

Have a go at finishing that bit.

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