1. $$ log_2(4) = $$

   $$ =log_2(2^2) = 2 $$

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2. $$ log_3(9) = $$

   $$ = log_3 (3^2) = 2$$

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3. $$ log_2 (32) = $$

   $$ = log_2(2^5) = 5 $$

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4. $$ log(1000) =$$

   $$ = log_{10} (1000)= $$

   $$ =log_{10}(10^3) =3 $$

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5. $$ log_2(0.8) = $$

   $$ =log_2 \left( \frac{8}{10} \right) =$$

   $$ = log_2 (8) - log_2(10) = $$

   $$ = log_2 (2^3) - log_2(2·5) = $$

   $$ = 3 -( log_2 (2·5) ) = $$

   $$ = 3 - (log_2(2) + log_2(5))= $$

   $$ = 3 - (1+log_2(5)) = $$

   $$ = 2 -log_2(5) $$

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6. $$ log_7(\sqrt{7}) = log_7(7^{\frac{1}{2}}) = $$

   $$ \frac{1}{2} \cdot log_7(7) = $$

   $$= \frac{1}{2}\cdot 1 $$

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7. 

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8. 

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9. 

We will use the logarithmic properties and that

$$log (1000) = log (10^3) = 3$$

So we can write the equation as

We have an equality between logarithms, so the arguments (what is inside) has to be the same:

The solution to the equation is x = 50.

We use the logarithmic properties and we write 3 as

to obtain an equality between logarithms.

The logarithms are worth the same when their arguments (what's inside) are the same

We resolve the equation:

Now we have to prove that for these values of x the arguments are not 0 nor negative.

Therefore, it is the solution.

In this equation, when applying the properties to obtain an equality of logarithms, we will need to resolve a quadratic equation:

Notice that the only possible solution is x = 3 due to the fact that the arguments have to be positive.

We will use that log(10) = 1:

$$ 2log(x)-log(x-6)=0 $$

$$ log(x^2) = log(x-6) $$

In order for the logarithms to be the same it is needed that

$$ x^2 = x-6 $$

We solve the quadratic equation:

$$ x^2 -x +6 = 0 $$

$$ x=\frac{1\pm \sqrt{1-24}}{2} $$

Because the discriminant is negative (-23), there are not any solutions (real ones).

Therefore, there is no solution.

As we have two terms that are the same, 3 raised to these terms will be the same number:

We raise to the power of 3 because 3 is the logarithm's base.

We have two terms that are the same, therefore, if we raise the base, 2, to these terms we have the same number

$$ 10^{log(x^2)} = 10^{-10} $$

$$ x^2 = 10^{-10} = \frac{1}{10^{10}} $$

Therefore, doing the square root:

$$ x = \pm \sqrt{\frac{1}{10^{10}}} = $$

$$ = \pm \frac{1}{10^5} $$

Because the argument of the logarithm is x squared, both values are solutions to the equation because when raising to the power of two both numbers are positive.

We will use that

$$ 3 = log_5(125) $$

Therefore,

We know that

$$ log(10) = 1$$

Therefore,

We can write 3 like a logarithm:

$$ log(1000) = 3 $$

Therefore,

The only solution is x = 1000 due to the fact that the argument of a logarithm can never be zero.

We write 4 like a logarithm:

$$ log(10000) = log(10^4) = 4 $$

Therefore

We equal the arguments (what's inside the logarithm) and we resolve the quadratic equation:

So, the arguments coincide when x = 3 and x = 2, therefore, the logarithms will be the same. But we have to prove that for these values of x the arguments are not 0 nor negative.

The only possible solution is x = 20, because the argument of a logarithm has to be necessarily positive.

The two square roots of the equation are solutions to the logarithmic equation because the two arguments are positive.

The only solution is

$$ x = \frac{12}{5} $$

because if x = 0, the argument of the denominator's logarithm is negative.

The only solution is x = 10.

We will apply the logarithmic properties and we will write 2 as

to obtain an equality between logarithms.

As we already have the equality of logarithms, they will have the same value when they have the same argument (what's inside), therefore, we obtain a quadratic equation:

Now we check that for these two (possible) solutions the arguments are positive:

Therefore, the only solution is x = 2.

We write 1 like \(1 = log(10)\). Therefore,

As we have an equality of logarithms (with the same base), we equal the arguments and resolve the equation:

Now we check that the arguments (the polynomials of the logarithms) are positive:

In this equation we have the unknown factor x in the base of the logarithm.

What we will do is to use the definition of logarithm, to calculate it.

In other words, we will use the following property:

$$ log_b (a) = c \Leftrightarrow b^c = a $$

Therefore,

We apply the change of variable

This way we obtain the following linear equation system

We resolve it and we undo the change of variable:

Notice that the first equation in the system is not logarithmic.

We isolate x in the first equation and we substitute in the second:

This system is similar to the one in the previous exercise:

We apply a change of variable:

So, we obtain:

The solution for the previous system is:

Now we undo the change of variable and we obtain the solutions to the initial system:

We isolate x in the first equation and we substitute in the second:

We apply the change of variable:

And we obtain the following system:

that has as solution:

Finally, we undo the change of variable:

We apply a change of variable:

what provides the following system:

that has as solution:

Finally, we undo the change of variable:

We apply the next change of variable:

So we obtain the following system:

that has as solution:

Finally, we undo the change of variable:

We do not need to check the solutions because they are positive.

First we apply the property logarithm of a product:

We apply the change of variable:

and we obtain the following system:

that has as solution:

Finally, we undo the change of variable:

We do not need to check the solutions.

We apply the property logarithm of a product:

Now we apply a change of variable:

And we obtain the following system:

that has as solution

Finally, we undo the change of variable:

We do not have to check the solutions.

We apply the properties logarithm of a quotient and logarithm of a power:

We apply the following change of variable:

and we obtain the system:

that has as solution

Finally, we undo the change of variable: