How do you solve $log (3 x + 1) = 1 + log (2 x - 1)$ $log (3x+1) = 1 + log (2x-1)$ ?

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How do you solve $log (3 x + 1) = 1 + log (2 x - 1)$ $log (3x+1) = 1 + log (2x-1)$ ?

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Answer 1 · 2015-12-23T01:41:44

The explanation is given below.

Explanation:

$log (3 x + 1) = 1 + log (2 x - 1)$ $log(3x+1)=1+log(2x-1)$
$log (3 x + 1) = log (10) + log (2 x - 1)$ $log(3x+1)=log(10)+log(2x-1)$ since $log (10) = 1$ $log(10)=1$
$log (3 x + 1) = log (10 (2 x - 1))$ $log(3x+1)=log(10(2x-1))$ Product rule of logarithms
$3 x + 1 = 10 (2 x - 1)$ $3x+1 = 10(2x-1)$ if logA = log B then A= B
$3 x + 1 = 20 x - 10$ $3x+1 = 20x-10$ distributive law

Now we have to solve for x using inverse operaitions to isolate x.

$3 x + 1 + 10 = 20 x - 10 + 10$ $3x+1 + 10 = 20x-10+10$

Adding 10 on both sides would remove $10$ $10$ from the right side of the equation

$3 x + 11 = 20 x$ $3x+11=20x$
$3 x + 11 - 3 x = 20 x - 3 x$ $3x+11-3x=20x-3x$

Subtracting $3 x$ $3x$ on both sides would isolate the term containing $x$ $x$ to one side of the equation.

$11 = 17 x$ $11=17x$

Dividing both sides by 17 we get

$\frac{11}{17} = \frac{17 x}{17}$ $11/17=(17x)/17$

$\frac{11}{17} = x$ $11/17 = x$

$x = \frac{11}{17}$ $x=11/17$ Answer

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How do you solve $log (3 x + 1) = 1 + log (2 x - 1)$ $log (3x+1) = 1 + log (2x-1)$ ?

1 Answer

Explanation:

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