Answer :
Let's solve each logarithmic equation one by one:
(a) [tex]\log_5(m-4) = \log_5(9)[/tex]
Since the logarithms on both sides have the same base, we can equate the arguments:
[tex]m - 4 = 9[/tex]
[tex]m = 13[/tex]
(b) [tex]\log(x) + \log(x+6)[/tex]
This expression can be combined using the logarithm property: [tex]\log(a) + \log(b) = \log(ab)[/tex]:
[tex]\log(x(x+6)) = \log(x^2 + 6x)[/tex]
There isn't an equation provided to solve here.
(c) [tex]\log_8(z+3) = \log_8(5)[/tex]
Since the logarithms on both sides have the same base, we can equate the arguments:
[tex]z + 3 = 5[/tex]
[tex]z = 2[/tex]
(d) [tex]\log_3(2x) - \log_3(x-14) = 2[/tex]
We can use the logarithm property: [tex]\log_b(a) - \log_b(b) = \log_b(\frac{a}{b})[/tex]:
[tex]\log_3 \left( \frac{2x}{x-14} \right) = 2[/tex]
By converting the logarithm to an exponential equation:
[tex]\frac{2x}{x-14} = 3^2[/tex]
[tex]\frac{2x}{x-14} = 9[/tex]
Cross-multiply to solve for [tex]x[/tex]:
[tex]2x = 9(x - 14)[/tex]
[tex]2x = 9x - 126[/tex]
[tex]7x = 126[/tex]
[tex]x = 18[/tex]
(e) [tex]\log_3(x) + \log_3(2x+5) = \log_3(12)[/tex]
Combine the logarithms on the left:
[tex]\log_3(x(2x+5)) = \log_3(12)[/tex]
[tex]\log_3(2x^2 + 5x) = \log_3(12)[/tex]
Equate the arguments:
[tex]2x^2 + 5x = 12[/tex]
[tex]2x^2 + 5x - 12 = 0[/tex]
Solve this quadratic equation by factoring or using the quadratic formula.
(f) [tex]\log_5(84) - \log_5(3y-4) = \log_5(2)[/tex]
Combine using the property [tex]\log_b(a) - \log_b(b) = \log_b(\frac{a}{b})[/tex]:
[tex]\log_5 \left( \frac{84}{3y-4} \right) = \log_5(2)[/tex]
Equate the arguments:
[tex]\frac{84}{3y-4} = 2[/tex]
Cross-multiply and solve for [tex]y[/tex]:
[tex]84 = 2(3y-4)[/tex]
[tex]84 = 6y - 8[/tex]
[tex]92 = 6y[/tex]
[tex]y = \frac{92}{6}[/tex]
[tex]y = \frac{46}{3}[/tex]
(g) [tex]\log_9(r) + \log_9(r+7) = \log_9(18)[/tex]
Combine the logarithms:
[tex]\log_9(r(r+7)) = \log_9(18)[/tex]
[tex]\log_9(r^2 + 7r) = \log_9(18)[/tex]
Equate the arguments:
[tex]r^2 + 7r = 18[/tex]
[tex]r^2 + 7r - 18 = 0[/tex]
Solve this quadratic equation [tex](r - 2)(r + 9) = 0[/tex]
The solutions are [tex]r = 2[/tex] or [tex]r = -9[/tex], but since [tex]\log_9(r)[/tex] must be defined, [tex]r[/tex] should be positive.
So, [tex]r = 2[/tex].
(h) [tex]\log(3n-5) = 3[/tex]
Convert the logarithm to an exponential equation:
[tex]3n - 5 = 10^3[/tex]
[tex]3n - 5 = 1000[/tex]
[tex]3n = 1005[/tex]
[tex]n = 335[/tex]
(i) [tex]\log_3(12m) - \log_3(1+m) = 2[/tex]
Combine the logarithms:
[tex]\log_3 \left( \frac{12m}{1+m} \right) = 2[/tex]
Convert to exponential form:
[tex]\frac{12m}{1+m} = 3^2[/tex]
[tex]\frac{12m}{1+m} = 9[/tex]
Cross-multiply:
[tex]12m = 9(1+m)[/tex]
[tex]12m = 9 + 9m[/tex]
[tex]3m = 9[/tex]
[tex]m = 3[/tex]
These solutions provide the values of the variables that satisfy the given logarithmic equations.