SOLUTION: Step-by-step solution of: {{{ 5^(2x+1) = 25^x*5^(3*x) }}}

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Question 1087279: Step-by-step solution of:
+5%5E%282x%2B1%29+=+25%5Ex%2A5%5E%283%2Ax%29+
Found 2 solutions by MathLover1, rothauserc:
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

+5%5E%282x%2B1%29+=+25%5Ex%2A5%5E%283%2Ax%29+
+5%5E%282x%2B1%29+=+%285%5E2%29%5Ex%2A5%5E%283x%29+
+5%5E%282x%2B1%29+=+5%5E%282x%29%2A5%5E%283x%29+
+5%5E%282x%2B1%29+=+5%5E%282x%2B3x%29+
+5%5E%282x%2B1%29+=+5%5E%285x%29+........if base same, exponents are same too
+2x%2B1+=5x+
+1+=5x-2x+
+1+=3x+
x=1%2F3



Answer by rothauserc(4718) About Me  (Show Source):
You can put this solution on YOUR website!
5^(2x+1) = 25^(x) * 5^(3x)
:
Note 25^(x) * 5^(3x) = (5^2)^(x) * 5^(3x) = 5^(2x) * 5^(3x) = 5^(5x)
:
5^(2x+1) = 5^(5x)
:
take logarithms of both sides of =
:
log(5) (2x+1) = 5log(5) (x)
:
divide both sides of = by log(5)
:
2x + 1 = 5x
:
3x = 1
:
*******
x = 1/3
*******
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