Question

I'm solving exponential and logarithmic equations and I have no idea what I'm doing. The equation is 8^(2x)=32

Answer 1

you write
\[8=2^3\] and so \[\large 8^{2x}=(2^3)^{2x}=2^{6x}\]

Answer 2

then since \(32=2^5\) you have
\[\large 2^{6x}=2^5\] making \(6x=5\) and you can solve for \(x\)

Answer 3

sat explained how to do it. The idea is IF you can write your equation to look like
\[ 2^{6x} = 2^5\]
then that means the exponents must be equal, i.e. 6x= 5
solving for x is "normal" algebra: divide both sides by 6 to find x= 5/6

So how do you change
\[ 8^{2x} = 32 \]
into that form ?

Answer 4

In these simple problems, look for both numbers (the 8 and the 32) to be powers of the same base. 2 divides into 8 to get 2*4. and 4 is 2*2, so 8= 2*2*2 or 2^3
your hope is 32 is also a power of 2. and it is, 32= 2^5