Answer :
Sure, let's solve the equation [tex]\(\log_7(3x^3 + x) - \log_7(x) = 2\)[/tex] step-by-step and identify any extraneous solutions.
First, let's use the properties of logarithms to combine the logarithms:
[tex]\[ \log_7(3x^3 + x) - \log_7(x) = \log_7\left(\frac{3x^3 + x}{x}\right) = \log_7(3x^2 + 1) \][/tex]
So, the equation simplifies to:
[tex]\[ \log_7(3x^2 + 1) = 2 \][/tex]
To clear the logarithm, rewrite the equation in exponential form:
[tex]\[ 3x^2 + 1 = 7^2 \][/tex]
Since [tex]\(7^2 = 49\)[/tex], the equation becomes:
[tex]\[ 3x^2 + 1 = 49 \][/tex]
Now, solve for [tex]\(x\)[/tex]:
[tex]\[ 3x^2 + 1 = 49 \][/tex]
[tex]\[ 3x^2 = 48 \][/tex]
[tex]\[ x^2 = 16 \][/tex]
[tex]\[ x = \pm 4 \][/tex]
So, the potential solutions are [tex]\(x = 4\)[/tex] and [tex]\(x = -4\)[/tex].
Next, we should check for extraneous solutions. A solution is extraneous if it does not satisfy the original equation or if it makes the argument of any logarithm non-positive (since logarithms of non-positive numbers are undefined).
Let's check both solutions:
1. For [tex]\(x = 4\)[/tex]:
[tex]\[ \log_7(3(4)^3 + 4) - \log_7(4) = 2 \][/tex]
[tex]\[ \log_7(3 \cdot 64 + 4) - \log_7(4) = 2 \][/tex]
[tex]\[ \log_7(192 + 4) - \log_7(4) = 2 \][/tex]
[tex]\[ \log_7(196) - \log_7(4) = 2 \][/tex]
[tex]\[ \log_7\left(\frac{196}{4}\right) = 2 \][/tex]
[tex]\[ \log_7(49) = 2 \][/tex]
Since [tex]\(7^2 = 49\)[/tex], this is true, so [tex]\(x = 4\)[/tex] is a valid solution.
2. For [tex]\(x = -4\)[/tex]:
Check the arguments of the logarithms:
[tex]\[ \log_7(3(-4)^3 + (-4)) - \log_7(-4) \][/tex]
[tex]\[ 3(-64) + (-4) = -192 - 4 = -196 \][/tex]
Since [tex]\(\log_7(-196)\)[/tex] and [tex]\(\log_7(-4)\)[/tex] are undefined (logarithms of negative numbers are undefined in the real number domain), [tex]\(x = -4\)[/tex] is not a valid solution. Therefore, [tex]\(x = -4\)[/tex] is extraneous.
Considering this, the extraneous solution to the given logarithmic equation is:
[tex]\[ x = -4 \][/tex]
First, let's use the properties of logarithms to combine the logarithms:
[tex]\[ \log_7(3x^3 + x) - \log_7(x) = \log_7\left(\frac{3x^3 + x}{x}\right) = \log_7(3x^2 + 1) \][/tex]
So, the equation simplifies to:
[tex]\[ \log_7(3x^2 + 1) = 2 \][/tex]
To clear the logarithm, rewrite the equation in exponential form:
[tex]\[ 3x^2 + 1 = 7^2 \][/tex]
Since [tex]\(7^2 = 49\)[/tex], the equation becomes:
[tex]\[ 3x^2 + 1 = 49 \][/tex]
Now, solve for [tex]\(x\)[/tex]:
[tex]\[ 3x^2 + 1 = 49 \][/tex]
[tex]\[ 3x^2 = 48 \][/tex]
[tex]\[ x^2 = 16 \][/tex]
[tex]\[ x = \pm 4 \][/tex]
So, the potential solutions are [tex]\(x = 4\)[/tex] and [tex]\(x = -4\)[/tex].
Next, we should check for extraneous solutions. A solution is extraneous if it does not satisfy the original equation or if it makes the argument of any logarithm non-positive (since logarithms of non-positive numbers are undefined).
Let's check both solutions:
1. For [tex]\(x = 4\)[/tex]:
[tex]\[ \log_7(3(4)^3 + 4) - \log_7(4) = 2 \][/tex]
[tex]\[ \log_7(3 \cdot 64 + 4) - \log_7(4) = 2 \][/tex]
[tex]\[ \log_7(192 + 4) - \log_7(4) = 2 \][/tex]
[tex]\[ \log_7(196) - \log_7(4) = 2 \][/tex]
[tex]\[ \log_7\left(\frac{196}{4}\right) = 2 \][/tex]
[tex]\[ \log_7(49) = 2 \][/tex]
Since [tex]\(7^2 = 49\)[/tex], this is true, so [tex]\(x = 4\)[/tex] is a valid solution.
2. For [tex]\(x = -4\)[/tex]:
Check the arguments of the logarithms:
[tex]\[ \log_7(3(-4)^3 + (-4)) - \log_7(-4) \][/tex]
[tex]\[ 3(-64) + (-4) = -192 - 4 = -196 \][/tex]
Since [tex]\(\log_7(-196)\)[/tex] and [tex]\(\log_7(-4)\)[/tex] are undefined (logarithms of negative numbers are undefined in the real number domain), [tex]\(x = -4\)[/tex] is not a valid solution. Therefore, [tex]\(x = -4\)[/tex] is extraneous.
Considering this, the extraneous solution to the given logarithmic equation is:
[tex]\[ x = -4 \][/tex]
