How do you solve #log (x + 3) = log (6) - log (2x-1)#?

Question

Theodore Amidon · Accepted Answer

# x = (-5 + sqrt(97))/4#

I'm assuming that all the #log# functions have the same basis #> 1#.

First of all, let's compute the domain of the logarithmic expressions. For the first one, #x + 3 > 0# must hold, so # x > -3#. For the last one, we get #2x -1 > 0 <=> x > 1/2#.

As the second condition is the more restrictive one, we can assume that the domain is #x > 1/2# and any possible solutions need to respect this condition.

Now, in order to "get rid" of the logarithmic expressions, it is necessary to simplify the terms on the right-hand side.

To do so, remember the logarithmic rule: #log_a(x) - log_a(y) = log_a(x/y)#

In our case, it means:

#log(x+3) = log(6) - log(2x-1)# #<=> log(x+3) = log(6/(2x-1))#

Now, due to the fact that for #x#, #y > 0# and #a != 1#, #log x = log y <=> x = y# holds, we can "drop" the logarithms on both sides of the equation.

#<=> x + 3 = 6/(2x-1)#

... multiply both sides with #2x-1#...

# <=> (x+3)(2x-1) = 6#

#<=> 2x^2 +5x - 9 = 0#

This is a quadratic equation that can be solved e. g. with the quadratic formula:

# x = (-b +- sqrt(b^2 - 4ac))/(2a)#

Here, #a = 2#, #b = 5#, #c = -9#, so we can compute the solution as follows:

#x = (-5 +- sqrt(25 - 4 * 2 * (-9))) / 4 = (-5 +- sqrt(97))/4#

The solution # x = (-5 - sqrt(97))/4# is negative, therefore we can ignore it because of our restriction to the domain #x > 1/2#.

So, we have just one solution: # x = (-5 + sqrt(97))/4 ~~ 1.2122#

Hope that this helped!

Serenity Jones · Answer

undefined To solve the equation $ \log(x + 3) = \log(6) - \log(2x - 1) $, you can use the properties of logarithms. First, apply the property $ \log(a) - \log(b) = \log\left(\frac{a}{b}\right) $ to combine the logarithms on the right side of the equation. Then, since the logarithm of the same base is equal to each other, you can drop the logarithm on both sides and solve for $ x $.

Here are the steps:

1. Apply the property $ \log(a) - \log(b) = \log\left(\frac{a}{b}\right) $:
$$ \log(x + 3) = \log\left(\frac{6}{2x - 1}\right) $$

2. Since the logarithm of the same base is equal, drop the logarithm on both sides:
$$ x + 3 = \frac{6}{2x - 1} $$

3. Multiply both sides of the equation by $ (2x - 1) $ to eliminate the fraction:
$$ (2x - 1)(x + 3) = 6 $$

4. Expand and simplify:
$$ 2x^2 + 6x - x - 3 = 6 $$
$$ 2x^2 + 5x - 3 - 6 = 0 $$
$$ 2x^2 + 5x - 9 = 0 $$

5. This is a quadratic equation. You can solve it using the quadratic formula:
$$ x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}} $$

Where $ a = 2 $, $ b = 5 $, and $ c = -9 $.

6. Plug in the values and solve for $ x $:
$$ x = \frac{{-5 \pm \sqrt{{5^2 - 4(2)(-9)}}}}{{2(2)}} $$
$$ x = \frac{{-5 \pm \sqrt{{25 + 72}}}}{{4}} $$
$$ x = \frac{{-5 \pm \sqrt{{97}}}}{{4}} $$

Therefore, the solutions for $ x $ are $ x = \frac{{-5 + \sqrt{97}}}{{4}} $ and $ x = \frac{{-5 - \sqrt{97}}}{{4}} $.