SOLUTION: These are exponential equations ( not that I know what the actually are); can someone please solve these before I die of stupidness!!!!
9^2x = 27^x - 1
5^n-1 = 1/ 25
25^x
Algebra.Com
Question 88931: These are exponential equations ( not that I know what the actually are); can someone please solve these before I die of stupidness!!!!
9^2x = 27^x - 1
5^n-1 = 1/ 25
25^x = 5^(x^2 - 15)
2^x * 4^x+5 = 4^2x-1
sqrt(3)^2x+4 = 9^x-2
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
9^2x = 27^(x - 1)
3^4x = 3^(3(x-1))
3^4x = 3^(3x-3)
Since the bases are the same the exponents must be equal:
4x = 3x-3
x = -3
-------------
5^n-1 = 1/ 25
5^(n-1) = 5^-2
n-1 = -2
n=-1
-----------
25^x = 5^(x^2 - 15)
5^2x = 5^(x^2-15)
2x = x^2-15
x^2-2x-15=0
(x-5)(x+3)=0
x=5 or x=-3
--------------
2^x * 4^x+5 = 4^2x-1
2^x * 2^(2x+10) = 2^(4x-2)
2^(3x+10) = 2^(4x-2)
3x+10 = 4x-2
x = 12
---------------
sqrt(3)^2x+4 = 9^x-2
3^[(1/2)(2x+4)] = 3^(3(x-2))
3^(x+2) = 3^(3x-6)
x+2 = 3x-6
2x=8
x=4
============
Be patient; it will come.
==============
Cheers,
Stan H.
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