SOLUTION: How do you solve: log5(3x-2)^4=8

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: How do you solve: log5(3x-2)^4=8       Log On


   



Question 1205441: How do you solve: log5(3x-2)^4=8

Found 5 solutions by math_tutor2020, greenestamps, ikleyn, Edwin McCravy, mccravyedwin:
Answer by math_tutor2020(3817) About Me  (Show Source):
You can put this solution on YOUR website!

If the 5 refers to base 5, then,
log%285%2C%28%283x-2%29%5E4%29%29+=+8

4%2Alog%285%2C%283x-2%29%29+=+8 Pull down the exponent using the rule log(A^B) = B*log(A)

log%285%2C%283x-2%29%29+=+8%2F4

log%285%2C%283x-2%29%29+=+2

3x-2+=+5%5E2 Convert to exponential form.

3x-2+=+25

3x+=+25%2B2

3x+=+27

x+=+27%2F3

x+=+9

Answer by greenestamps(13203) About Me  (Show Source):
You can put this solution on YOUR website!


log%285%2C%283x-2%29%5E4%29=8

Convert to exponential form:

%283x-2%29%5E4=5%5E8
Take the 4th root of both expressions:

3x-2=5%5E2=25
3x=27
x=9

ANSWER: x=9


Answer by ikleyn(52855) About Me  (Show Source):
You can put this solution on YOUR website!
.
How do you solve: log5(3x-2)^4=8
~~~~~~~~~~~~~~~~~~~~~


        It is really interesting and educative problem,  because it has a  HUGE  underwater stone.

        But its formulation in the post is  INCORRECT.

        A correct formulation is   "find all real solutions to this equation   log%285%2C%28%283x-2%29%5E4%29%29 = 8".


Notice that left side of the equation is determined for all real values of x except 2/3.

Again, it is not a typo: left side of this equation is determined for all real values of x, 
positive and negative, except of x = 2/3.


If you write it in exponential form, you will get

    %283x-2%29%5E4 = 5%5E8.    (1)


If you take square root of both sides of (1), you will get

    %283x-2%29%5E2 = 5%5E4 = 625.    (2)


In (2), left side is non-negative as a square - therefore, in the right side we write non-negative value  625  of the square root  sqrt%285%5E8%29.


If you take square root of both sides of (2), you will get

    3x-2 = +/-sqrt%28625%29 = +/- 25.    (3)


Left side of (3) can be positive or negative - therefore, we leave two possibility for the right side to be +25 or -25.


Equation (3) has two solutions.

One solution   is  3x-2 = 25,   3x = 25+2 = 27,    x = 27/3 = 9.

Other solution is  3x-2 = -25,  3x = -25+2 = -23,  x = -23/3.


ANSWER.  The given equation has 2 (two, TWO) solutions in real numbers.

         They are x= 9  and  x= -23/3.

-----------------------

At this point,  the problem is solved in full.
Two real solutions are possible, and they both are found.
You can check it on your own that both numbers  9  and  -23/3  are the solutions to the given equation.


Two other tutors lost one of the two possible solutions in real numbers.
In  Math  (in entering exams to some universities,  for example),  it is considered as a grave sin.


The cost of such mistake is losing the entrance exam in  Math.



Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!
Ikleyn has pointed out that there is something to watch for in
the "Pulling down the exponent" rule, as math_tutor2020 calls it.

That's when the exponent is an even integer and its base contains a 
variable.

However, the "pull down" rule can still be used without losing an answer,
without complicating the problem by switching to exponents so soon, by
inserting %22%22+%2B-+%22%22 in front of the base of the even number exponent 
when you "pull down the exponent".

log%285%2C%28%283x-2%29%5E4%29%29%22%22=%22%228

4%2Alog%285%2C%28%22%22+%2B-+%283x-2%29%5E%22%22%29%29%22%22=%22%228

log%285%2C%28%22%22+%2B-+%283x-2%29%5E%22%22%29%29%22%22=%22%222

%22%22+%2B-+%283x-2%29%22%22=%22%225%5E2

3x-2%22%22=%22%22%22%22+%2B-+25

3x-2 = 25    3x-2 = -25
  3x = 27      3x = -23
   x = 9        x = -23/3

Edwin

Answer by mccravyedwin(408) About Me  (Show Source):
You can put this solution on YOUR website!

It has occurred to me that 

X%5E%282n%29%5E%22%22 and %28%22%22+%2B-+X%5E%22%22%29%5E%282n%29 mean the same for any positive integer n.

So for any positive even integer 2n, we can always replace X%5E%282n%29%5E%22%22 with %28%22%22+%2B-+X%5E%22%22%29%5E%282n%29. 
So this problem should make use of this substitution at the very beginning.

------------------------------------------

log%285%2C%28%283x-2%29%5E4%29%29%22%22=%22%228

log%285%2C%28%28%22%22+%2B-+%283x-2%29%5E%22%22%29%5E4%29%29%22%22=%22%228

4%2Alog%285%2C%28%22%22+%2B-+%283x-2%29%5E%22%22%29%29%22%22=%22%228

log%285%2C%28%22%22+%2B-+%283x-2%29%5E%22%22%29%29%22%22=%22%222

%22%22+%2B-+%283x-2%29%22%22=%22%225%5E2

3x-2%22%22=%22%22%22%22+%2B-+25

3x-2 = 25    3x-2 = -25
  3x = 27      3x = -23
   x = 9        x = -23/3

Edwin