nmomma83
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(Need asap)Solve each equation. Use one of the 4 methods you have practiced the last few days:
1. Write exponents using the same base
2. Take the log of both sides
3. Use properties of logs then write as an exponent
4. Use properties of logs then drop the log on both sides

11. Log5 6+log5 2x^2=log5 48

Respuesta :

sqdancefan

Answer:

  x = -2 or +2

Step-by-step explanation:

You want to solve the logarithmic equation ...

  [tex]\log_5(6)+\log_5(2x^2)=\log_5(48)[/tex]

Properties of logs

The terms on the left can be combined using the property of logarithms:

  log(a)+log(b) = log(ab)

This gives you ...

  [tex]\log_5(6\cdot2x^2)=\log_5(48)[/tex]

Now, we can drop the logs on both sides:

  12x² = 48

Quadratic

This can be solved in the usual way:

  x² = 4 . . . . . . divide by 12

  x = ±√4 = ±2 . . . . . take the square root

The solutions are x = -2 or +2.

msm555

Answer:

[tex]x = 2[/tex] and [tex]x = - 2[/tex]

Step-by-step explanation:

To solve the equation [tex]\log_5 6 + \log_5 (2x^2) = \log_5 48[/tex], we can use logarithmic properties to condense the left side into a single logarithm.

First, recall the product rule of logarithms which states that:

[tex]\Large\boxed{\boxed{\sf \log_b (M) + \log_b (N) = \log_b (MN)}}[/tex].

Applying this rule to your equation:

[tex]\log_5 (6) + \log_5 (2x^2) = \log_5 (6 \cdot 2x^2)[/tex]

Simplify [tex]6 \cdot 2x^2[/tex] to [tex]12x^2[/tex]:

[tex]\log_5 (12x^2) = \log_5 48[/tex]

Now that both sides of the equation are in the same base, we can equate their arguments:

[tex]12x^2 = 48[/tex]

Divide both sides by 12:

[tex]x^2 = \dfrac{48}{12}[/tex]

[tex]x^2 = 4[/tex]

Now, take the square root of both sides:

[tex]x = \pm \sqrt{4}[/tex]

[tex]x = \pm 2[/tex]

So, the solutions to the equation are [tex]x = 2[/tex] and [tex]x = -2[/tex].

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