Write the expression as a sum and/or difference of logarithms. Express powers as factors. ln⁡2x1+8x(x−3)13,x&gt;3 ln⁡2x1+8x(x−3)13= (Simplify your answer.)

Step 1 The given expression is ln⁡2x1+8x(x−3)13,x&gt;3 Step 2 Using the formula ln⁡(ab)=ln⁡a−ln⁡b ln⁡(ab)=ln⁡a+ln⁡b ln⁡an=nln⁡a Step 3 On simplifying with the help of given formula ln⁡2x1+8x(x−3)13=ln⁡2x(1+8x)12(x−3)13 =ln⁡2x(1+8x)12−ln⁡(x−3)13 =ln⁡2x+ln⁡(1+8x)12−ln⁡(x−3)13 =ln⁡2x+ln⁡(1+8x)12−ln⁡(x−3)13 =ln⁡2+ln⁡x+12ln⁡(1+8x)−13ln⁡(x−3) ⇒ln⁡2x1+8x(x−3)13=ln⁡2+ln⁡x+12ln⁡(1+8x)−13ln⁡(x−3)

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vazelinahS

Answered question

2021-08-03

Write the expression as a sum and/or difference of logarithms. Express powers as factors.

\ln \frac{2 x \sqrt{1 + 8 x}}{{(x - 3)}^{13}}, x > 3

\ln \frac{2 x \sqrt{1 + 8 x}}{{(x - 3)}^{13}} =

(Simplify your answer.)

Answer & Explanation

Bella

Skilled2021-08-04Added 81 answers

Step 1
The given expression is

\ln \frac{2 x \sqrt{1 + 8 x}}{{(x - 3)}^{13}}, x > 3

Step 2
Using the formula

\ln (\frac{a}{b}) = \ln a - \ln b

\ln (a b) = \ln a + \ln b

{\ln a}^{n} = n \ln a

Step 3
On simplifying with the help of given formula

\ln \frac{2 x \sqrt{1 + 8 x}}{{(x - 3)}^{13}} = \ln \frac{2 x {(1 + 8 x)}^{\frac{1}{2}}}{{(x - 3)}^{13}}

= \ln 2 x {(1 + 8 x)}^{\frac{1}{2}} - {\ln (x - 3)}^{13}

= \ln 2 x + {\ln (1 + 8 x)}^{\frac{1}{2}} - {\ln (x - 3)}^{13}

= \ln 2 x + {\ln (1 + 8 x)}^{\frac{1}{2}} - {\ln (x - 3)}^{13}

= \ln 2 + \ln x + \frac{1}{2} \ln (1 + 8 x) - 13 \ln (x - 3)

\Rightarrow \ln \frac{2 x \sqrt{1 + 8 x}}{{(x - 3)}^{13}} = \ln 2 + \ln x + \frac{1}{2} \ln (1 + 8 x) - 13 \ln (x - 3)

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