SOLUTION: I just need someone to explain this and why 2 is the answer: 3^(log base 9 (4))= 2 I'm not sure how to write out log base correctly, but it's log(4) with the base of 9.

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: I just need someone to explain this and why 2 is the answer: 3^(log base 9 (4))= 2 I'm not sure how to write out log base correctly, but it's log(4) with the base of 9.      Log On


   

Question 1057131: I just need someone to explain this and why 2 is the answer:
3^(log base 9 (4))= 2
I'm not sure how to write out log base correctly, but it's log(4) with the base of 9.
Found 4 solutions by solve_for_x, josgarithmetic, ikleyn, MathTherapy:
Answer by solve_for_x(190) About Me  (Show Source):
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When you have something like log (base 9) 4, you can rewrite it as log+4+%2F+log+9

And when you do that, you can use any base for the logarithms that you want.

So, instead of using the common base-10 log, we can write is using a base-3 log:

log9+%284%29+=+log+4+%2F+log+9+=+log3+%284%29+%2F+log3+%289%29

But log3 (9) = 2:

log9+%284%29+=+log3+%284%29+%2F+2

Substituting log3%284%29%2F2 in place of log9%284%29 gives:


3%5E%28log3%284%29%2F2%29

which equals:

sqrt%283%5E%28log3%284%29%29%29++

And since 3%5E%28log3%284%29%29+=+4+, the expression is equal to:

sqrt%284%29+=+2

Answer by josgarithmetic(39618) About Me  (Show Source):
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3^(log(9,4))=2

3%5E%28log%289%2C4%29%29=2 ?

Look into that log%289%2C4%29. Change of Base formula might be helpful....


log%2810%2C4%29%2Flog%2810%2C9%29=log%289%2C4%29

Look in table of logs or use calculator:
log%289%2C4%29=%280.60206%29%2F%280.9542425%29=0.63093

3%5E0.63093, and you might use logs table again...
Result is something extremely close to 2.

Answer by ikleyn(52790) About Me  (Show Source):
You can put this solution on YOUR website!
.
I just need someone to explain this and why 2 is the answer:
3^(log base 9 (4))= 2
I'm not sure how to write out log base correctly, but it's log(4) with the base of 9.
~~~~~~~~~~~~~~~~~~~~~~~

The problem is solved in two steps: 

1.  log%289%2C%284%29%29 = log%283%2C%282%29%29


    Indeed, if log%283%2C%282%29%29 = x, it means (by the definition of logarithm) that 3%5Ex = 2.
    Then  %283%5Ex%29%5E2 = 2%5E2 = 4, which implies  3%5E%282x%29 = 4, which is the same as %283%5E2%29%5Ex = 4,  or  9%5Ex = 4.
    The last equality means that log%289%2C%284%29%29 = x.  Hence,  log%289%2C%284%29%29 = log%283%2C%282%29%29,  QED.



2.  3%5E%28log%283%2C%282%29%29%29 = 2.    (<---- b%5E%28log%28b%2C%28x%29%29%29 = x  for any x > 0 and for any b > 0,  b=/= 1.  
                                 This is the basic property of the logarithm, equivalent to the definition of the logarithm)

On logarithms, see the lessons
    - WHAT IS the logarithm
    - Properties of the logarithm
in this site.

Also, you have this free of charge online textbook in ALGEBRA-I in this site
    - ALGEBRA-I - YOUR ONLINE TEXTBOOK.

The referred lessons are the part of this online textbook under the topic "Logarithms".


Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!
I just need someone to explain this and why 2 is the answer:
3^(log base 9 (4))= 2
I'm not sure how to write out log base correctly, but it's log(4) with the base of 9.
3%5E%28log+%289%2C+%284%29%29%29

Let 3%5E%28log+%289%2C+%284%29%29%29+=+x

log+%283%2C+%28x%29%29+=+log+%289%2C+%284%29%29 ------- Converting from EXPONENTIAL to LOGARITHMIC form

log+%283%2C+%28x%29%29+=+log+%283%2C+%284%29%29%2Flog%283%2C+%289%29%29 ------- Changing right side to base 3, by applying CHANGE OF BASE

log+%283%2C+%28x%29%29+=+log+%283%2C+%284%29%29%2F2 ------- Changing log+%283%2C+%289%29%29 to 2

2+%2A+log%283%2C+%28x%29%29+=+log+%283%2C+%284%29%29 ----- Cross-multiplying

log+%283%2C+%28x%29%5E2%29+=+log+%283%2C+%284%29%29 ------ Applying a+%2A+log+%28b%2C+%28c%29%29+=+log+%28b%2C%28c%29%5Ea%29 to left-side of equation

x%5E2+=+4 ------ Equating expressions since log bases are the same