Question 924766
Solve using exponent laws:

a) 9^4x-1=27(81^x-1)
If the exponent of 9 is 4x-1, and the exponent of 81 is x-1:
9 = 3^2
27 = 3^3
--> {{{3^(2*(4x-1)) = 27*(3^(4*(x-1)))}}}
{{{3^(8x-2) = 3^3*(3^(4x-4))}}}
{{{3^(8x-2) = (3^(4x-1))}}}
8x-2 = 4x-1
4x = 1
x = 1/4
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b) 7^2x+1= index of 3radical sign with 49 inside of it
If you mean:
{{{7^(2x+1) = root(3,49)}}}
{{{7^(2x+1) = 7^(2/3)}}}
2x+1 = 2/3
x = -1/6
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Put parentheses around the exponents to eliminate ambiguity.  They're free.