Solve the equation ( 2^x - 1 )*[log 2 (2x)] = 0

Expert Answers
hala718 eNotes educator| Certified Educator

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

since we have a product of two functions. the:

(2^x - 1) = 0 

==>: 2^x = 1

==> x= 0  (impossible because the function is not defined for x=0)

OR:

log 2 (2x) = 0  

==> 2x = 2^0 = 1

==> x = 1/2

neela | Student

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.

giorgiana1976 | Student

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

thewriter | Student

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.

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