SOLUTION: how am i supposed to solve this: Solve: 1/9=27^(4x-3)

Question 307744: how am i supposed to solve this:
Solve: 1/9=27^(4x-3)
Answer by rapaljer(4671) (Show Source): You can put this solution on YOUR website!
1/9=27^(4x-3)

Notice that the base numbers here (that is, 1/9 and 27) are both powers of 3. This makes this a very special problem that can be solved in a special way.

1/9=3^-2
27=3^3

By law of logarithms, when you raise a power to a power, you multiply exponents.

Since the base numbers are the same, the powers are equal:
-2=12x-9

Add +9 to each side:
-2+9=12x
7=12x

I have quite a bit of explanation with examples solved in LIVING COLOR on my own website. Please go to "www2.seminolestate.edu/rrapalje". Look for the links on my Homepage "Basic, Intermediate, and College Algebra: One Step at a Time" and also "MATH IN LIVING COLOR". You will find specific problems of this type in "College Algebra" (Chapter 4), Section 4.04 Solving Exponential and Logarithmic Equations.

I also have an entire explanation of Logarithms on video from my own teaching in class before I retired a few years ago. To see the videos, look for "Rapalje Videos in Living Color", and see College Algebra for the topic of Logarithms.

I think you will find the explanations, both written and video, much easier to understand than traditional textbooks. My own students did!!

Dr. Robert J. Rapalje, Retired
Seminole State College of Florida
Altamonte Springs Campus

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