How do you solve #log (x + 3) = log (6)  log (2x1)#?
Now, in order to "get rid" of the logarithmic expressions, it is necessary to simplify the terms on the righthand side.
In our case, it means:
This is a quadratic equation that can be solved e. g. with the quadratic formula:
Hope that this helped!
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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:

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

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

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

Expand and simplify: [ 2x^2 + 6x  x  3 = 6 ] [ 2x^2 + 5x  3  6 = 0 ] [ 2x^2 + 5x  9 = 0 ]

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 ).
 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}} ).
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When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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