Solve Log5 (126 - z) + log5 z = 3

Next, we need to remove the logarithm by rewriting the equation in exponential form. Since the base of the logarithm is 5, we rewrite the equation as:

[tex]126z - z^2 = 5^3[/tex]

Calculate [tex]5^3[/tex]:

[tex]5^3 = 125[/tex]

Substitute this back into the equation:

[tex]126z - z^2 = 125[/tex]

Rearrange it to form a quadratic equation:

[tex]-z^2 + 126z - 125 = 0[/tex]

To make it easier to solve, multiply through by -1:

[tex]z^2 - 126z + 125 = 0[/tex]

Now, we will solve this quadratic equation using the quadratic formula:

[tex]z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}[/tex]

Here, [tex]a = 1[/tex], [tex]b = -126[/tex], and [tex]c = 125[/tex]. Substitute these values into the formula:

[tex]z = \frac{-(-126) \pm \sqrt{(-126)^2 - 4 \cdot 1 \cdot 125}}{2 \cdot 1}[/tex]

[tex]z = \frac{126 \pm \sqrt{15876 - 500}}{2}[/tex]

[tex]z = \frac{126 \pm \sqrt{15376}}{2}[/tex]

Calculate [tex]\sqrt{15376}[/tex]:

[tex]\sqrt{15376} = 124[/tex] (since 124 is the integer square root)

Substitute back to find the solutions for [tex]z[/tex]:

[tex]z = \frac{126 + 124}{2}[/tex] or [tex]z = \frac{126 - 124}{2}[/tex]

[tex]z = \frac{250}{2} = 125[/tex] or [tex]z = \frac{2}{2} = 1[/tex]

Thus, the solutions for [tex]z[/tex] are [tex]z = 125[/tex] and [tex]z = 1[/tex].

Finally, check both solutions in the context of the original logarithmic equation.

For [tex]z = 125[/tex]: [tex]\log_5 (126 - 125) + \log_5 125 = \log_5 1 + \log_5 125 = 0 + \log_5 125 = 3[/tex].
For [tex]z = 1[/tex]: [tex]\log_5 (126 - 1) + \log_5 1 = \log_5 125 + 0 = 3[/tex].

No extraneous solutions, so both [tex]z = 125[/tex] and [tex]z = 1[/tex] are valid solutions.