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

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 ${\log}_{10} \left(x\right) = 2$. The answer to that is 100.

Rewrite the given question as:

${\log}_{10} \left(x + 6\right) + {\log}_{10} \left(3 x - 2\right) = {\log}_{10} \left(100\right) \ldots \ldots \ldots \left(1\right)$

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

Consider the LHS only

${\log}_{10} \left(x + 6\right) + {\log}_{10} \left(3 x - 2\right) \to {\log}_{10} \left[\textcolor{w h i t e}{.} \left(x + 5\right) \left(3 x - 2\right) \textcolor{w h i t e}{.}\right]$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Rewrite equation (1) as

${\log}_{10} \left[\textcolor{w h i t e}{.} \left(x + 5\right) \left(3 x - 2\right) \textcolor{w h i t e}{.}\right] = {\log}_{10} \left(200\right)$

$\implies \left(x + 5\right) \left(3 x - 2\right) = 200$

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

Have a go at finishing that bit.