Solving Log Equations?

Question

Solve each log equation below.

(1) log_5 (x) = 3

(2) log_3 (3x - 2) = 2

Answer 1

Logs can be rewritten as log_b c= a --> b^a=c

log_5 (x) = 3 can be rewritten as 5^3=x
so x= 125

log_3 (3x - 2) = 2 can be rewritten as 3^2=3x-2
3^2=3x-2
9=3x-2
11=3x
11/3=x

Answer 2

Since the left side is the sum of two logs, then it is the log of a product: log (x + 2)(x - 1) = log 28 log (x² + x - 2) = log 28. Set x² + x - 2 = 28, and solve for x by factoring: x² + x - 2 = 28 x² + x - 2 - 28 = 0 x² + x - 30 = 0 (x + 6)(x - 5) = 0 x + 6 = 0 → x = -6 x - 5 = 0 → x = 5. Now we have to do a little thinking. If we let x = -6, then (x + 2) = -6 + 2 = -4, and (x - 1) = -6 - 1 = -7. Since we can't take the logs of negative numbers, then x = -6 is not a possible solution. Thus x = 5 is the only viable solution. Now we check to make sure x = 5 works in the original equation: log (5 + 2) + log (5 - 1) = log 28 log 7 + log 4 = log 28 log (7•4) = log 28 log 28 = log 28. There is another check we can perform. That is by actually adding the logs of 7 and 4, then checking to see if their sum approximates the log of 28: log 7 ≈ 0.8451 log 4 ≈ 0.6021 log 28 ≈ 1.4472 Thus log 7 + log 4 ≈ 1.4472 ≈ log 28.

Answer 3

(1) I take it that it is Log to base 5 of x is 3 and I need to find x

x = 5^3 = 125.

(2) Log to base 3 of (3x-2) is 2. Therefore, 3x-2=3^2

3x-2=9, 3x=11, therefore x=11/3 or 3.666666666..............

Answer 4

x = 5^3 = 125
3x-2 = 3^2 = 9
3x = 11
x = 11/3

Answer 5

log5(x) =3

5^3 =x

log3(3x-2)=2
3^2=3x-2
9 = 3x-2
11/3 = x

Answer 6

solve your own math problems...
THINK :)

rather write it as:

log (5*x) = 3
log (3*(3*x-2)) = 2

* PS, use LN both sides of the equation to "cancel" the LOG