# Solve for "a" in terms of "r" when `theta` = 45degrees and v=3w : lnr=lna-ln(cos`theta` )-(v/w)ln(sec`theta` +tan`theta` ) Leave the answer in exact form.

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To solve for in `a` terms of `r,`

Substitute `theta = 45` and `v = 3w`

`ln r = ln a - ln (cos 45) - ((3w)/w) ln(sec 45 + tan 45)`

Simplify,

`ln r = ln a - ln (sqrt(2)/2) - 3 ln (2/sqrt(2) + 1)`

Then use the Property of Natural Logarithm,

`ln r = ln a - ln (sqrt(2)/2) - ln (sqrt(2) + 1)^3`

Factor,

`ln r = ln a - [ln (sqrt(2)/2) + ln (sqrt(2) + 1)^3]`

Use again the Property of Natural Logarithm,

`lnr=ln(a /((sqrt(2)/2)(sqrt(2)+1)^3))`

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Raise each side by `e,`

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`e^(lnr)=e^(ln(a /((sqrt(2)/2)(sqrt(2)+1)^3)))`

` ` `r=a/((sqrt(2)/2)(sqrt(2)+1)^3)`

Expand `(sqrt(2)+1)^3`

`r=a/((sqrt(2)/2)(5 sqrt(2)+7))`

Multiply the denominator,

`r=a/(5+7 sqrt(2)/2)`

``Apply cross-multiplication,

`a = r (5 + 7 sqrt(2)/2)`

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Given

ln(r)=ln(a)-ln(cos )-(v/w)ln(sec +tan )

where θ = 45 degrees and v/w = 3

ln(a) = ln(r) + ln(cosθ) +3ln(secθ+tanθ)

= ln(rcosθ)+3ln (secθ+tanθ)

=ln(rcosθ) +ln(secθ+tanθ)^3

=ln(rcosθ).(secθ+tanθ)^3

`:.` a= (rcosθ) (secθ+tanθ)^3

=(r/**√2**)(1+**√2**)^3

=(r/**√2**)(1+2**√2**+3**√2**+6)

=(r/**√2)**(7+5**√2)**

** a =(r/2****) (7****√2+10) -Ans**

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