3 to the power of x = 81 to the power of 2x solve the equation?

Answer 1

There are a few ways of approaching this problem.
1. Involving logarithms
By log(n)x I will mean log with the base of n of x
We will use the rule that log(n)x^a=a*log(n)x
let's log your equation with base 3
log(3)3^x=log(3)81^2x
using the rule
x*log(3)3 =2x*log(3)81
now, we know that 81=9*9=3*3*3*3=3^4
so log(3)81=4*log(3)3
also, log(n)n=1
thus we get x=2*x*4=8x
so, simplifying 7x=0
x=0
which was an obvious answer in the first place because x^0=1 for positive xs

Now, we might not want to use logarithms openly (we will use them anyway, just discreetly.
We can use the fact that a^(x*y)=(a^x)^y
so, we have 3^x=81^(2x)
but we've established, that 81=3^4
so 3^x=(3^4)^(2x)
using the fact above
3^x=3^(4*2x)
so
3^x=3^(8x)
now we can argument that when two exponential expressions are equal and their bases are equal, the exponents must be equal as well. (here we use the logarithm covertly)
thus x=8x =>x=0
There are also other solutions in the complex number domain...
Hope I've helped

Answer 2

3^x = 81^2x
or, 3^x = 3^6x {81 = 3^3}
or, x = 6x
=> x = 0

Answer 3

3^(x) = (3^(4))^(2x)
3^(x) = 3^(8x)
x = 8x
x = 0