Is 2sinh(ln(x))+(1^2)/2-4/3 =2 ?

The answer to whether 2sinh(ln(x))+(1^2)/2-4/3 =2 is False

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prove 2sinh(ln(x))+(1^2)/2-4/3 =2

Solution

prove

2 sinh (ln (x)) + \frac{1 ^{2}}{2} - \frac{4}{3} = 2

Solution

False

Solution steps

2 sinh (ln (x)) + \frac{1 ^{2}}{2} - \frac{4}{3} = 2

Show that the two sides are not equal

For

2 sinh (ln (x)) + \frac{1 ^{2}}{2} - \frac{4}{3} = 2

plug in

x = 1

2 sinh (ln (1)) + \frac{1 ^{2}}{2} - \frac{4}{3} = - \frac{5}{6} (Decimal : - 0.83333 \dots)

2 sinh (ln (1)) + \frac{1 ^{2}}{2} - \frac{4}{3}

Simplify:

ln (1) = 0

ln (1)

Apply log rule:

lo g_{a} (1) = 0

= 0

2 sinh (0) + \frac{1 ^{2}}{2} - \frac{4}{3} = \frac{12 sinh ( 0 ) - 5}{6}

2 sinh (0) + \frac{1 ^{2}}{2} - \frac{4}{3}

Apply rule

1^{a} = 1

1^{2} = 1

= 2 sinh (0) + \frac{1}{2} - \frac{4}{3}

Convert element to fraction:

2 sinh (0) = \frac{2 sinh ( 0 )}{1}

= \frac{2 sinh ( 0 )}{1} + \frac{1}{2} - \frac{4}{3}

Least Common Multiplier of

1, 2, 3 : 6

1, 2, 3

Least Common Multiplier (LCM)

Prime factorization of

1

Prime factorization of

2 : 2

2

2

is a prime number, therefore no factorization is possible

= 2

Prime factorization of

3 : 3

3

3

is a prime number, therefore no factorization is possible

= 3

Compute a number comprised of factors that appear in at least one of the following:

1, 2, 3

= 2 \cdot 3

Multiply the numbers:

2 \cdot 3 = 6

= 6

Adjust Fractions based on the LCM

Multiply each numerator by the same amount needed to multiply its

corresponding denominator to turn it into the LCM

6

For

\frac{2 sinh ( 0 )}{1} :

multiply the denominator and numerator by

6

\frac{2 sinh ( 0 )}{1} = \frac{2 sinh ( 0 ) \cdot 6}{1 \cdot 6} = \frac{12 sinh ( 0 )}{6}

For

\frac{1}{2} :

multiply the denominator and numerator by

3

\frac{1}{2} = \frac{1 \cdot 3}{2 \cdot 3} = \frac{3}{6}

For

\frac{4}{3} :

multiply the denominator and numerator by

2

\frac{4}{3} = \frac{4 \cdot 2}{3 \cdot 2} = \frac{8}{6}

= \frac{12 sinh ( 0 )}{6} + \frac{3}{6} - \frac{8}{6}

Since the denominators are equal, combine the fractions:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{12 sinh ( 0 ) + 3 - 8}{6}

Add/Subtract the numbers:

3 - 8 = - 5

= \frac{12 sinh ( 0 ) - 5}{6}

= \frac{12 sinh ( 0 ) - 5}{6}

Rewrite using trig identities:

sinh (0) = 0

sinh (0)

Use the Hyperbolic identity:

sinh (x) = \frac{e ^{x} - e ^{- x}}{2}

= \frac{e ^{0} - e ^{- 0}}{2}

Simplify

= 0

= \frac{12 \cdot 0 - 5}{6}

Simplify

= - \frac{5}{6}

The two sides are not equal

\Rightarrow False

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Is 2sinh(ln(x))+(1^2)/2-4/3 =2 ?
The answer to whether 2sinh(ln(x))+(1^2)/2-4/3 =2 is False