SOLUTION: 32^(4x-1) = 16^(3x) Trying to help my son with his algebra 11 homework. I have no idea about this one. Help!

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: 32^(4x-1) = 16^(3x) Trying to help my son with his algebra 11 homework. I have no idea about this one. Help!      Log On


   

Question 733342: 32^(4x-1) = 16^(3x)
Trying to help my son with his algebra 11 homework. I have no idea about this one. Help!
Found 2 solutions by josmiceli, Edwin McCravy:
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
+32%5E%28+4x-1+%29+=+16%5E%283x%29+ is this it? If so,
Note that +32+=+2%5E5+ and +16+=+2%5E4+
Now write this as:
+%28+2%5E5+%29%5E%28+4x-1+%29+=+%28+2%5E4+%29%5E%28+3x+%29+
-------------------------------
Now use the general rule:
+%28a%5Eb%29%5Ec++=+a%5E%28b%2Ac%29+
+2%5E%28+5%2A%28+4x-1+%29+%29+=+2%5E%28+4%2A3x+%29+
For the equation to be true, the exponents on
both sides have to be equal, so
+5%2A%28+4x-1+%29+=+12x+
+20x+-+5+=+12x+
+8x+=+5+
+x+=+5%2F8+
------------
check:
+32%5E%28+4x-1+%29+=+16%5E%283x%29+
+32%5E%28+4%2A%285%2F8%29+-+1+%29+=+16%5E%283%2A%285%2F8%29%29+
+32%5E%28+5%2F2+-+1+%29+=+16%5E%28+15%2F8+%29+
+32%5E%28+3%2F2+%29+=+16%5E%28+15%2F8+%29+
square both sides
+32%5E3+=+16%5E%28+15%2F4+%29+
+32%5E3+=+2%5E15+
+32768+=+32768+
OK

Answer by Edwin McCravy(20055) About Me  (Show Source):
You can put this solution on YOUR website!
324x-1 = 163x

32 = 4·8 = (2·2)·(2·4) = (2·2)·(2·2·2) = 2·2·2·2·2 = 25

16 = 4·4 = (2·2)·(2·2) = 2·2·2·2 = 24

So we replace 32 with (25) and 16 with (24):

(25)4x-1 = (24)3x

We remove the parentheses by multiplying the exponents:

220x-5 = 212x

Since the bases are the same on both sides, and they are positive and
not equal to 1, we can set the exponents equal to each other:

20x - 5 = 12x
     8x = 5
      x = 5%2F8

Edwin