What is the answer to x+log_{2}(3-2^x)=1 ?

The answer to x+log_{2}(3-2^x)=1 is x=0,x=1

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

x+log_{2}(3-2^x)=1

x + lo g_{2} (3 - 2^{x}) = 1

x = 0, x = 1

Solution steps

x + lo g_{2} (3 - 2^{x}) = 1

Move

x

to the right side

lo g_{2} (3 - 2^{x}) = 1 - x

f (x) = g (x),

then

a^{f (x)} = a^{g (x)}

2^{l o g_{2} (3 - 2^{x})} = 2^{1 - x}

Simplify

2^{l o g_{2} (3 - 2^{x})} : 3 - 2^{x}

3 - 2^{x} = 2^{1 - x}

Solve

3 - 2^{x} = 2^{1 - x} : x = 0, x = 1

x = 0, x = 1

Verify Solutions:

x = 0

True

, x = 1

True

The solutions are

x = 0, x = 1

lo g_{9} (a) = 2

2 = lo g_{3} (t)

lo g_{2} (3 x + 4) = 4

lo g_{x} (2) = - \frac{2}{3}

7^{x \cdot l o g_{7} (x - 2)} = x - 2

What is the answer to x+log_{2}(3-2^x)=1 ?
The answer to x+log_{2}(3-2^x)=1 is x=0,x=1
Solve for x: x+log_{2}(3-2^x)=1
The solution is x=0,x=1