Respuesta :

alinakincsem

Answer:

The correct answer option is x = 4.

Step-by-step explanation:

We are given the following logarithmic equation and we are to determine whether which of the given options shows its extraneous solution:

[tex] log _ 7 ( 3 x ^ 3 + x ) - log _ 7 ( x ) = 2 [/tex]

We can rewrite it as:

[tex]log7[\frac{3x^3+x}{x} ]=2[/tex]

But we know that [tex]log_7(49)=2[/tex]

So, [tex]log7[\frac{3x^3+x}{x} ]=log_7(49)[/tex]

Cancelling the log to get:

[tex]\frac{3x^3+x}{x} =49[/tex]

Further simplifying it to get:

[tex]3x^2+1=49[/tex]

[tex]3x^2=48[/tex]

[tex]x^2=\frac{48}{3}[/tex]

[tex]x^2=16[/tex]

x = 4

natagutty

Answer:

The extraneous solution to the logarithmic equation is [tex]x=-4[/tex]

Step-by-step explanation:

We have the equation:

[tex]Log_{7} (3x^3+x)-Log_7(x)=2[/tex]

By properties of logarithms:

[tex]LogA-LogB=Log(\frac{A}{B})[/tex]

So, with the equation we have:

[tex]Log_{7} \frac{(3x^3+x)}{x}=2[/tex]

[tex]Log_{7}( \frac{3x^3+x}{x})=2\\Log_{7}( \frac{3x^3}{x}+\frac{x}{x})=2\\Log_{7}( \frac{3x^3}{x}+1)=2\\Log_{7}(3x^2+1)=2[/tex]

This logarithm base is 7 and this equation is equal to 2,  the number 7 passes as the base on the other side of the equation and the two as an exponent, after that we just to find x:

[tex]7^2=(3x^2+1)\\49=3x^2+1\\49-1=3x^2\\\frac{48}{3} =x^2\\16=x^2[/tex]

Now, we can find x with square root

[tex]16=x^2\\\sqrt{16} =\sqrt{x^2} \\x_1=4\\x_2=-4[/tex]

This equation has two answers because it is a quadratic equation, so with this logic the strange solution is -4

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