SOLUTION: What is the solution to this problem? 32^2y = 5^(4y+1)

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: What is the solution to this problem? 32^2y = 5^(4y+1)      Log On


   

Question 1135343: What is the solution to this problem?
32^2y = 5^(4y+1)
Found 3 solutions by Alan3354, MathLover1, rothauserc:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
If it's 32^(2y) = 5^(4y+1):
2y*log(32) = (4y+1)*log(5) = 4y*log(5) + log(5)
2y*log(32) - 4y*log(5) = log(5)
y*(2log(32) - 4log(5)) = log(5)
y = log(5)/(log(1024 - log(625))
y = log(5)/log(1024/625)
y =~ 3.259818
====================
PS 32^2y = 5^(4y+1) is not a problem, it's an equation.
"Solve for y" would be a problem.
========
Yes, I am picky and pedantic.

Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

32%5E%282y%29+=+5%5E%284y%2B1%29

log%2832%5E%282y%29%29+=+log%285%5E%284y%2B1%29%29

2y%2Alog%2832%29+=+%284y%2B1%29log%285%29.........simplify log%2832%29+=log%282%5E5%29+=5log%282%29

10y%2Alog%282%29+=+4y%2Alog%285%29%2Blog%285%29
10y%2Alog%282%29+-+4y%2Alog%285%29=log%285%29
y%2810log%282%29+-+4log%285%29%29=log%285%29
y%28log%282%5E10%29+-+log%285%5E4%29%29=log%285%29
y%2Alog%28%282%5E10%2F5%5E4%29%29=log%285%29
y=log%285%29%2Flog%28%281024%2F625%29%29
y=log%285%29%2Flog%28%2832%5E2%2F25%5E2%29%29
y=log%285%29%2F2log%28%2832%2F25%29%29

Answer by rothauserc(4718) About Me  (Show Source):
You can put this solution on YOUR website!
32^(2y) = 5^(4y+1)
:
(32^2)^y = (5^(4)^y) * 5
:
1024^y = 625^y * 5
:
take ln (natural log) of both sides of =
:
yln(1024) = yln(625) +ln(5)
:
Note ln(a^b) = bln(a) and ln(ab) = ln(a) + ln(b)
:
yln(1024) - yln(625) = ln(5)
:
y(ln(1024) - ln(625)) = ln(5)
:
yln(1024/625) = ln(5)
:
Note ln(a) - ln(b) = ln(a/b)
:
y = ln(5)/ln(1024/625)
:
************
y = 3.2598
************
: