(3/5 )^-5 × (5/3 )^11 = ( 5/3)^8x , then the value of x is
Answers
Answer:
Integrate the function
\huge\green\tt\frac{ \sqrt{tanx} }{sinxcosx}}
⇛\huge\tt\frac{ \sqrt{tanx} }{sinxcosx}sinxcosxtanx
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⇛\huge\tt \frac{ \sqrt{tanx} }{sinxcosx \times \frac{cosx}{cosx}}sinxcosx×cosxcosxtanx
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⇛\huge\tt \frac{ \sqrt{tanx} }{sinx \times \frac{ {cos}^{2} x}{cosx}}sinx×cosxcos2xtanx ㅤ ㅤ ㅤ
⇛ \huge\tt\frac{ \sqrt{tanx} }{ {cos}^{2} x \times \frac{sinx}{cosx} }cos2x×cosxsinxtanx
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⇛\huge\tt\frac{ \sqrt{tanx} }{ {cos}^{2}x \times tanx }cos2x×tanxtanx
⇛\huge\tt {tan}^{ \frac{1}{2} - 1 } \times \frac{1}{ {cos}^{2} x}tan21−1×cos2x1 ㅤ ㅤ ㅤ ㅤ ㅤ
⇛\huge\tt {(tan)}^{ - \frac{ 1}{2} } \times \frac{1}{ {cos}^{2}x } = {(tanx)}^{ - \frac{1}{2} } \times {sec}^{2} x⇛(tan)(tan)−21×cos2x1=(tanx)−21×sec2x⇛(tan)
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⇛\huge\tt {(tan)}^{ - \frac{ 1}{2} } \times \frac{1}{ {cos}^{2}x } = ∫ {(tanx)}^{ - \frac{1}{2} } \times {sec}^{2} x \times dx⇛(tan)(tan)−21×cos2x1=∫(tanx)−21×sec2x×dx⇛(tan)
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\bold\blue{☛\: Let tanx=t}☛Lettanx=t
\bold\blue{☛ \:Differentiating \: both \: sides \: w.r.t.x}☛Differentiatingbothsidesw.r.t.x
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⇛\huge\tt {sec}^{2} x = \frac{dt}{dx}sec2x=dxdt
⇛\huge\tt{dx \frac{dt}{ {sec}^{2}x }}dxsec2xdt
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⇛\huge\tt∴∫ {(tanx)}^{ - \frac{1}{2} } \times {sec}^{2} x \times dx∴∫(tanx)−21×sec2x×dx
⇛\huge\tt ∫ {(t)}^{ - \frac{1}{2} } \times {sec}^{2} x \times \frac{dt}{ {sec}^{2}x }∫(t)−21×sec2x×sec2xdt
⇛\huge\tt ∫ {t}^{ - \frac{1}{2} }∫t−21 ㅤ ㅤ
⇛ \huge\tt\frac{ {t}^{ - \frac{1}{2} + 1} }{ - \frac{1}{2} + 1 }−21+1t−21+1
⇛ \huge\tt \frac{ {t}^{ \frac{1}{2} } }{ \frac{1}{2} } + c = 2 {t}^{ \frac{1}{2} } + c = 2 \sqrt{t}21t21+c=2t21+c=2t
⇛\huge2 \sqrt{t} + c = 2 \sqrt{tanx}2t+c=2tanx
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Answer:
Step-by-step explanation:
(5/3) ^-5 * (5/3)^11 = (5/3) ^ 8x, then x =?
=(5/3) ^- 5 +11 = (5/3)^8x
= (5/3) ^6 = (5/3)^8x
On equation the exponents
6 =8x
=> x = 6/8 = 3/4 (ans).