What is the answer to solvefor j,y= 1/((2^j)) ?

The answer to solvefor j,y= 1/((2^j)) is j=-(ln(y))/(ln(2))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

solvefor j,y= 1/((2^j))

solve for

j, y = \frac{1}{( 2 ^{j} )}

j = - \frac{ln ( y )}{ln ( 2 )}

Solution steps

y = \frac{1}{( 2 ^{j} )}

Apply exponent rules

ln (y) = - j ln (2)

Solve

ln (y) = - j ln (2) : j = - \frac{ln ( y )}{ln ( 2 )}

j = - \frac{ln ( y )}{ln ( 2 )}

Verify Solutions:

j = - \frac{ln ( y )}{ln ( 2 )}

True

The solution is

j = - \frac{ln ( y )}{ln ( 2 )}

2 \cdot 3^{x} + 1 = 10. 3^{12} \cdot 8. 3^{12}

9 = 3^{m}

\frac{2 ^{x} + 1}{( 2 ^{x} )} = \frac{41}{4}

2^{x + 1} + 2^{x + 2} = 48

3.34 = 3.2 1^{n}

What is the answer to solvefor j,y= 1/((2^j)) ?
The answer to solvefor j,y= 1/((2^j)) is j=-(ln(y))/(ln(2))