What is ln((1+2x)^{1/2})*1/2 (1+2x)^{-1/2} ?

The solution to ln((1+2x)^{1/2})*1/2 (1+2x)^{-1/2} is (ln(2x+1))/(4(2x+1)^{1/2)}

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simplify ln((1+2x)^{1/2})*1/2 (1+2x)^{-1/2}

simplify

ln ((1 + 2 x)^{\frac{1}{2}}) \cdot \frac{1}{2} (1 + 2 x)^{- \frac{1}{2}}

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

Solution steps

ln ((1 + 2 x)^{\frac{1}{2}}) \frac{1}{2} (1 + 2 x)^{- \frac{1}{2}}

Apply exponent rule:

a^{- b} = \frac{1}{a ^{b}}

= \frac{1}{2} \cdot \frac{1}{( 2 x + 1 ) ^{\frac{1}{2}}} ln ((2 x + 1)^{\frac{1}{2}})

Simplify

ln ((1 + 2 x)^{\frac{1}{2}}) : \frac{1}{2} ln (2 x + 1)

= \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{( 2 x + 1 ) ^{\frac{1}{2}}} ln (2 x + 1)

Multiply fractions:

a \cdot \frac{b}{c} \cdot \frac{d}{e} = \frac{a \cdot b \cdot d}{c \cdot e}

= \frac{1 \cdot 1 \cdot 1 \cdot ln ( 1 + 2 x )}{2 \cdot 2 ( 1 + 2 x ) ^{\frac{1}{2}}}

Refine

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

k

e^{- 1 l n (5 x)}

L t e^{- 2 t}

(9.11 \cdot 1 0^{- 31}) (3 \cdot 1 0^{8})

What is ln((1+2x)^{1/2})*1/2 (1+2x)^{-1/2} ?
The solution to ln((1+2x)^{1/2})*1/2 (1+2x)^{-1/2} is (ln(2x+1))/(4(2x+1)^{1/2)}