Nash Frank

Answered

2022-07-17

a)Express the given quantity as a single logarithm.

$\frac{1}{3}\mathrm{ln}(x+2{)}^{3}+\frac{1}{2}[\mathrm{ln}(x)-\mathrm{ln}({x}^{2}+3x+2{)}^{2}]$

b) Solve each equation for x.

1)${e}^{5-4x}=4$

x=?

2) $\mathrm{ln}(3x-13)=8$

x=?

$\frac{1}{3}\mathrm{ln}(x+2{)}^{3}+\frac{1}{2}[\mathrm{ln}(x)-\mathrm{ln}({x}^{2}+3x+2{)}^{2}]$

b) Solve each equation for x.

1)${e}^{5-4x}=4$

x=?

2) $\mathrm{ln}(3x-13)=8$

x=?

Answer & Explanation

Jeroronryca

Expert

2022-07-18Added 13 answers

a)Solution:

$\frac{1}{3}\mathrm{ln}(x+2{)}^{3}+\frac{1}{2}[\mathrm{ln}(x)-\mathrm{ln}({x}^{2}+3x+2{)}^{2}]$

$=\mathrm{ln}(x+2{)}^{3/3}+\frac{1}{2}\mathrm{ln}x-\frac{1}{2}\mathrm{ln}({x}^{2}+3x+2{)}^{2}$

$=\mathrm{ln}(x+2)+\mathrm{ln}{x}^{1/2}-\mathrm{ln}({x}^{2}+3x+2{)}^{2/2}$

$=\mathrm{ln}(x+2)({x}^{1/2})-\mathrm{ln}({x}^{2}+3x+2)$

${[\because \mathrm{ln}a+\mathrm{ln}b=\mathrm{ln}ab]}$

$=\mathrm{ln}\frac{({x}^{3/2}+2{x}^{1/2})}{({x}^{2}+3x+2)}$

${[\because \mathrm{ln}a-\mathrm{ln}b=\mathrm{ln}\frac{a}{b}]}$

${\text{Hence}\text{}\frac{1}{3}\mathrm{ln}(x+2{)}^{3}+\frac{1}{2}[\mathrm{ln}x-\mathrm{ln}({x}^{2}+3x+2{)}^{2}]}$

${=\mathrm{ln}\frac{({x}^{3/2}+2{x}^{1/2})}{({x}^{2}+3x+2)}}$

$\frac{1}{3}\mathrm{ln}(x+2{)}^{3}+\frac{1}{2}[\mathrm{ln}(x)-\mathrm{ln}({x}^{2}+3x+2{)}^{2}]$

$=\mathrm{ln}(x+2{)}^{3/3}+\frac{1}{2}\mathrm{ln}x-\frac{1}{2}\mathrm{ln}({x}^{2}+3x+2{)}^{2}$

$=\mathrm{ln}(x+2)+\mathrm{ln}{x}^{1/2}-\mathrm{ln}({x}^{2}+3x+2{)}^{2/2}$

$=\mathrm{ln}(x+2)({x}^{1/2})-\mathrm{ln}({x}^{2}+3x+2)$

${[\because \mathrm{ln}a+\mathrm{ln}b=\mathrm{ln}ab]}$

$=\mathrm{ln}\frac{({x}^{3/2}+2{x}^{1/2})}{({x}^{2}+3x+2)}$

${[\because \mathrm{ln}a-\mathrm{ln}b=\mathrm{ln}\frac{a}{b}]}$

${\text{Hence}\text{}\frac{1}{3}\mathrm{ln}(x+2{)}^{3}+\frac{1}{2}[\mathrm{ln}x-\mathrm{ln}({x}^{2}+3x+2{)}^{2}]}$

${=\mathrm{ln}\frac{({x}^{3/2}+2{x}^{1/2})}{({x}^{2}+3x+2)}}$

Darryl English

Expert

2022-07-19Added 2 answers

b)Solution:

1)${e}^{5-4x}=4$

$\Rightarrow 5-4x=\mathrm{ln}4$

$\Rightarrow 4x=5-\mathrm{ln}4$

$\Rightarrow x=\frac{5}{4}-\frac{1}{4}\mathrm{ln}4$

2)$\mathrm{ln}(3x-13)=8$

$\Rightarrow 3x-13={e}^{8}$

$\Rightarrow 3x=13+{e}^{8}$

math xmlns="http://www.w3.org/1998/Math/MathML">⇒ x = 1 3 ( 13 + e 8 )

1)${e}^{5-4x}=4$

$\Rightarrow 5-4x=\mathrm{ln}4$

$\Rightarrow 4x=5-\mathrm{ln}4$

$\Rightarrow x=\frac{5}{4}-\frac{1}{4}\mathrm{ln}4$

2)$\mathrm{ln}(3x-13)=8$

$\Rightarrow 3x-13={e}^{8}$

$\Rightarrow 3x=13+{e}^{8}$

math xmlns="http://www.w3.org/1998/Math/MathML">

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