To solve (2^x-1)*[log2 (2x)} = 0

Solution:

(2^x-1)*{log2 (2x) } = 0

By zero product rule, if ab = 0, the a =0 or b =0.

So 2^x-1 = 0 , Or log2 (2x) = 0.

2^x -1 = givex 2^x = 1 = 2^0. So base are equal. So exponents x = 0.

log2(2x) = 0 gives :

log2(2) + log2 (x) = 0

log2 (x) = - log2 (2) = (-1) log2 (2)

log2 (x) = log2 ( 2^-1) , as m*log a = log a^m.

x = 2^-1

x = 1/2.

First, before solving the equation, we'll impose the constraint of existence of logarithm:

2x>0

x>0

Now, we'll re-write the term log 2 (2x), using the product property of logarithms:

log 2 (2x) = log 2 2 + log 2 x = 1 + log 2 x

We'll re-write the entire equation:

( 2^x - 1 )*(1 + log 2 x) = 0

We'll set each factor as zero:

2^x - 1 = 0

2^x = 1

2^x = 2^0

**x = 0**

1 + log 2 x = 0

log 2 x = -1

x = 2^-1

**x = 1/2**

**Since the solutions of the equation have to be positive, but different from zero, the eq. will have only one solution, **

**x = 1/2**

We have to solve the equation (2^x-1)*[log 2(2x)]=0

So either (2^x-1)=0 or [log 2(2x)]=0

If (2^x-1)=0, 2^x=1 or x=0

and if [log 2(2x)]=0, 2(2x)=1 or x=1/4

Of the two solutions x=0 is not acceptable as log 0 is not defined.

Therefore **x=1/4 **is the only acceptable answer.