Answer:
see attached
Step-by-step explanation:
We prefer to solve these by rewriting the equations to the form f(x) = 0, then having a graphing calculator show us the x-intercepts of f(x). This is illustrated in the second attachment.
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The applicable rules of logarithms are ...
- log(a) +log(b) = log(ab) . . . for logs of any base
- log(a) = b ⇔ 10^b = a
- ln(e^a) = a
- log(a) = log(b) ⇔ a = b . . . . for a > 0 and b > 0
We can apply these rules to the given expressions to solve for x algebraically.
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log(x-1) ...
Taking antilogs, we have ...
(x -1)(5x) = 10^2 = 100
x(x -1) = 20 . . . . . . . . . . . divide by 5
x² -x -20 = 0 . . . . subtract 20, put in standard form
(x -5)(x +4) = 0 . . . . factor
x = 5 or x = -4 . . . . . . the latter is an extraneous solution
x = 5 only
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ln(x +5) ...
Taking antilogs, we have ...
x +5 = (x +1)(x -1)
x² -x -6 = 0 . . . . . . . subtract (x+5), write in standard form
(x -3)(x +2) = 0 . . . . factor
x = 3 or x = -2 . . . . . . the latter is an extraneous solution
x = 3 only
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e^x² ...
Taking natural logarithms, we have ...
x² = 4x +5
x² -4x -5 = 0 . . . . . write in standard form
(x -5)(x +1) = 0 . . . . factor
x = 5 or x = -1 . . . . values that make the factors zero
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log₄(5x² ...
Taking antilogs, we have ...
5x² +2 = x +8
5x² -x -6 = 0 . . . . . write in standard form
(5x -6)(x +1) = 0 . . . . factor
x = 6/5 or x = -1 . . . . values that make the factors zero
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Additional comment
A solution is extraneous when it does not satisfy the original equation. Here, solutions are extraneous because they make the argument of the log function be negative in the original equation. The log function is not defined for negative arguments. (Actually, it gives complex values for negative arguments.)
We could have gone to the trouble to determine the applicable domain of each of the log equations. It is easier to (a) use a graphing calculator, or (b) test the solutions found.
