(2^4)(4^8)(8^16)(16^32)=32^x. Find x?

Answer 1

Firefly's approach is good. If you haven't seen logarithms yet, you can still solve the problem by writing everything as a power of 2, i.e.

4^8 = (2^2)^8 = 2^(2*8) = 2^16
8^16 = (2^3)^16 = 2^(3*16) = 2^48
16^32 = (2^4)^32 = 2^(4*32) = 2^128
32^x = (2^5)^x = 2^(5x)

So, rewrite your equation as (2^4)(2^16)(2^48)(2^128) = 2^5x, or 2^(4+16+48+128) = 2^5x. This reduces to 4+16+48+128 = 5x, or x = 196/5.

Answer 2

first let's take log2 of each side

log2(abcd) = log2(a)+log(2(b)+log2(c)+log2(d) so the left side of the equation will use this relationship, and the equation becomes
log2(2^4) + log2(4^8) + log2(8^16)+log2(16^32) = log2(32^x)
= xlog2(32)
now, log2(2^4) is easy. It's 4*log2(2) = 4
The rest goes like this
4+8log2(4)+16log2(8)+32log2(16) = xlog2(32)

or
4+8log2(2^2)+16log2(2^3)+32log2(2^4)=xlog2(2^5)

or
4+8*2*log2(2) + 16*3*log2(2) + 32*4*log2(2) = x*5*log2(2)

or
4+16+48+128 = 5x
or
x = (4+16+48+128)/5