Please help me solve A) 27^y-1 = 9^2y - 4?

Question

and

B) 2^x = 4^x - 1

Thank you

Answer 1

A) 27^y-1 = 9^2y - 4?

27 equals 3^3 and 9 equals 3^2 therefore the equation is equal to:
3^3(y-1)=3^2(2y-4)
because the bases are the same the exponents are equal.
3y-3=4y-8
then just solve:
y=5
B) 2^x = 4^x - 1
you use the same method here.
2 is 2^1 and 4 is 2^2
2^x=2^2(x-1)
x=2x-2
x=2

Answer 2

3^3y-1=3^4y-4 if you call 3^y=z
z^3-1=z^4-4
z^4-z^3-3=0 Thats supposing that -1 and -4 are not in the exponent
z=1.6581 (with a calculator) so ylog 3 = log 1.6581 and
y= log1.6581/log 3
If the -1 and -4 are in the exponent the problem becomes
z^3/27 =z^4/9^4
so z= 9^4/27 = 243 and y=log243/log3
B 2^x=2^2x-1
2^x=z
z^2-z-1=0 z= ((1+sqrt(5))/2 z must be >0
Than take log

Answer 3

Hint: Use logarithms

Edit: I'm trying to solve for ya, but is it 27^(y-1) or 27^y-1 (exponent y-1 or just y)
Clarify plz =)

Answer 4

its , x^4-x^3-3 = 0 ,

i've got two zeroes ,
one above -1.5 and below -1
another above 1.5 and below 2