SOLUTION: If (3^5x)(27^5x)=(3^48)(9^9x), then what is the value of x?
I tried to solve for x as a regular equation and got stuck when x was isolated on one side as an exponent.....(3^5x)(
Algebra.Com
Question 629636: If (3^5x)(27^5x)=(3^48)(9^9x), then what is the value of x?
I tried to solve for x as a regular equation and got stuck when x was isolated on one side as an exponent.....(3^5x)(27^5x)=(3^48)(9^8x)
81^10x=(3^48)(9^8x)
9^2x=(3^48)
Also tried....
10x=48+8x
2x=48
x=24
the answer is 12
Thank you
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
(3^5x)(27^5x)=(3^48)(9^8x)
(3^5x)((3^3)^5x)=(3^48)((3^2)^8x)
(3^5x)(3^(15x))=(3^48)(3^(16x))
3^(5x+15x) = 3^(48+16x)
3^(20x) = 3^(48+16x)
Since the bases are equal, the exponents are equal. So
20x = 48+16x
20x - 16x = 48
4x = 48
x = 48/4
x = 12
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