Logaritmische formules herleiden
Herleid tot de gevraagde vorm.
3p
Schrijf de formule \(y = 170 x^{-1{,}47}\) in de vorm \(\log(y) = a + b ⋅ \log(x) \text{.}\)
Geef \(a\) in twee decimalen.
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\(y = 170 x^{-1{,}47}\)
\(\log(y) = \log(170 x^{-1{,}47})\)
1p
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\(\log(y) = \log(170) + \log(x^{-1{,}47})\)
\(\log(y) = \log(170) - 1{,}47 ⋅ \log(x)\)
1p
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\(\log(y) = 2{,}230... - 1{,}47 ⋅ \log(x)\)
Dus \(y = 2{,}23 - 1{,}47 ⋅ \log(x) \text{.}\)
1p
Herleid tot de gevraagde vorm.
3p
Schrijf de formule \(y = {740 \over x^{2}}\) in de vorm \(\log(y) = a + b ⋅ \log(x) \text{.}\)
Geef \(a\) in twee decimalen.
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\(y = {740 \over x^{2}} = 740 x^{-2}\)
\(\log(y) = \log(740 x^{-2})\)
1p
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\(\log(y) = \log(740) + \log(x^{-2})\)
\(\log(y) = \log(740) - 2 ⋅ \log(x)\)
1p
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\(\log(y) = 2{,}869... - 2 ⋅ \log(x)\)
Dus \(y = 2{,}87 - 2 ⋅ \log(x) \text{.}\)
1p
Herleid tot de gevraagde vorm.
3p
Schrijf de formule \(\log(y) = 2{,}91 - 1{,}41 ⋅ \log(x)\) in de vorm \(y = a x^{b} \text{.}\)
Geef \(a\) in gehelen.
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\(\log(y) = 2{,}91 - 1{,}41 ⋅ \log(x)\)
\(\log(y) = \log(10^{2{,}91}) + \log(x^{-1{,}41})\)
\(\log(y) = \log(10^{2{,}91} ⋅ x^{-1{,}41})\)
1p
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\(y = 10^{2{,}91} ⋅ x^{-1{,}41}\)
1p
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\(y = 812{,}830... ⋅ x^{-1{,}41}\)
Dus \(y = 813 ⋅ x^{-1{,}41} \text{.}\)
1p
Herleid tot de gevraagde vorm.
3p
Schrijf de formule \(y = 6\,200 ⋅ 0{,}9^{x}\) in de vorm \(\log(y) = a x + b \text{.}\)
Geef \(a\) in vier decimalen en \(b\) in twee decimalen.
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\(y = 6\,200 ⋅ 0{,}9^{x}\)
\(\log(y) = \log(6\,200 ⋅ 0{,}9^{x})\)
\(\log(y) = \log(6\,200) + \log(0{,}9^{x})\)
1p
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\(\log(y) = \log(6\,200) + x ⋅ \log(0{,}9)\)
1p
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\(\log(y) = 3{,}792... + x ⋅ -0{,}04575...\)
Dus \(\log(y) = -0{,}0458 x + 3{,}79\)
1p
Herleid tot de gevraagde vorm.
3p
Schrijf de formule \(y = 7\,100 ⋅ 0{,}89^{5 x + 3}\) in de vorm \(\log(y) = a x + b \text{.}\)
Geef \(a\) in vier decimalen en \(b\) in twee decimalen.
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\(y = 7\,100 ⋅ 0{,}89^{5 x + 3}\)
\(\log(y) = \log(7\,100 ⋅ 0{,}89^{5 x + 3})\)
\(\log(y) = \log(7\,100) + \log(0{,}89^{5 x + 3})\)
1p
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\(\log(y) = \log(7\,100) + (5 x + 3) ⋅ \log(0{,}89)\)
\(\log(y) = \log(7\,100) + 5 x ⋅ \log(0{,}89) + 3 ⋅ \log(0{,}89)\)
1p
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\(\log(y) = 3{,}851... + 5 x ⋅ -0{,}05060... + 3 ⋅ -0{,}05060...\)
\(\log(y) = 3{,}851... - 0{,}25304... ⋅ x - 0{,}15182...\)
Dus \(\log(y) = -0{,}2530 x + 3{,}70\)
1p
Herleid tot de gevraagde vorm.
3p
Schrijf de formule \(\log(y) = -0{,}3593 x + 2{,}99\) in de vorm \(y = b ⋅ g^{x} \text{.}\)
Geef \(b\) in gehelen en \(g\) in twee decimalen.
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\(\log(y) = -0{,}3593 x + 2{,}99\)
\(y = 10^{-0{,}3593 x + 2{,}99}\)
1p
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\(y = 10^{-0{,}3593 x} ⋅ 10^{2{,}99}\)
\(y = (10^{-0{,}3593})^{x} ⋅ 10^{2{,}99}\)
1p
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\(y = 0{,}437...^{x} ⋅ 977{,}237...\)
Dus \(y = 977 ⋅ 0{,}44^{x} \text{.}\)
1p
Herleid tot de gevraagde vorm.
3p
Schrijf de formule \(y = 2{,}55 ⋅ {}^{4}\!\log(x) - 1{,}52\) in de vorm \(y = {}^{4}\!\log(a x^{b}) \text{.}\)
Geef \(a\) en \(b\) in twee decimalen.
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\(y = 2{,}55 ⋅ {}^{4}\!\log(x) - 1{,}52\)
\(\text{ } = {}^{4}\!\log(x^{2{,}55}) - 1{,}52\)
1p
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\(\text{ } = {}^{4}\!\log(x^{2{,}55}) + {}^{4}\!\log(4^{-1{,}52})\)
\(\text{ } = {}^{4}\!\log(x^{2{,}55} ⋅ 4^{-1{,}52})\)
1p
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\(\text{ } = {}^{4}\!\log(x^{2{,}55} ⋅ 0{,}121...)\)
Dus \(y = {}^{4}\!\log(0{,}12 ⋅ x^{2{,}55}) \text{.}\)
1p
Herleid tot de gevraagde vorm.
3p
Schrijf de formule \(y = {}^{3}\!\log({55 \over x^{4} \sqrt{x}})\) in de vorm \(y = a + b ⋅ {}^{3}\!\log(x) \text{.}\)
Geef \(a\) in twee decimalen.
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\(y = {}^{3}\!\log({55 \over x^{4} \sqrt{x}})\)
\(\text{ } = {}^{3}\!\log(55 x^{-4{,}5})\)
1p
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\(\text{ } = {}^{3}\!\log(55) + {}^{3}\!\log(x^{-4{,}5})\)
\(\text{ } = {}^{3}\!\log(55) - 4{,}5 ⋅ {}^{3}\!\log(x)\)
1p
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\(\text{ } = 3{,}647... - 4{,}5 ⋅ {}^{3}\!\log(x)\)
Dus \(y = 3{,}65 - 4{,}5 ⋅ {}^{3}\!\log(x) \text{.}\)
1p
Herleid tot de gevraagde vorm.
3p
Schrijf de formule \(y = {}^{3}\!\log(1{,}5 x) - 2{,}7\) in de vorm \(y = a + b ⋅ {}^{5}\!\log(x) \text{.}\)
Geef \(a\) en \(b\) in twee decimalen.
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\(y = {}^{3}\!\log(1{,}5 x) - 2{,}7\)
\(\text{ } = {}^{3}\!\log(1{,}5) + {}^{3}\!\log(x) - 2{,}7\)
1p
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\(\text{ } = {}^{3}\!\log(1{,}5) - 2{,}7 + {{}^{5}\!\log(x) \over {}^{5}\!\log(3)}\)
\(\text{ } = {}^{3}\!\log(1{,}5) - 2{,}7 + {1 \over {}^{5}\!\log(3)} ⋅ {}^{5}\!\log(x)\)
1p
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\(\text{ } = 0{,}369... - 2{,}7 + {1 \over 0{,}682...} ⋅ {}^{5}\!\log(x)\)
\(\text{ } = -2{,}330... + 1{,}464... ⋅ {}^{5}\!\log(x)\)
Dus \(y = -2{,}33 + 1{,}46 ⋅ {}^{5}\!\log(x) \text{.}\)
1p
Herleid tot de gevraagde vorm.
3p
Schrijf de formule \(y = 6 ⋅ {}^{2}\!\log(48 x) + 7\) in de vorm \(y = a + b ⋅ {}^{2}\!\log(3 x) \text{.}\)
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\(y = 6 ⋅ {}^{2}\!\log(48 x) + 7\)
\(\text{ } = 6 ⋅ ({}^{2}\!\log(16) + {}^{2}\!\log(3 x)) + 7\)
1p
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\(\text{ } = 6 ⋅ (4 + {}^{2}\!\log(3 x)) + 7\)
1p
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\(\text{ } = 24 + 6 ⋅ {}^{2}\!\log(3 x) + 7\)
\(\text{ } = 31 + 6 ⋅ {}^{2}\!\log(3 x)\)
1p
Druk \(x\) uit in \(y \text{.}\)
3p
\(y = 18 + 3 ⋅ {}^{5}\!\log(2 x - 4)\)
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\(y = 18 + 3 ⋅ {}^{5}\!\log(2 x - 4)\)
\(3 ⋅ {}^{5}\!\log(2 x - 4) = y - 18\)
\({}^{5}\!\log(2 x - 4) = \frac{1}{3} y - 6\)
1p
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\(2 x - 4 = 5^{\frac{1}{3} y - 6}\)
1p
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\(2 x = 5^{\frac{1}{3} y - 6} + 4\)
\(x = \frac{1}{2} ⋅ 5^{\frac{1}{3} y - 6} + 2\)
1p