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Linear Logarithmic Equations & y = ax n. Exponentia l form. Suppose that y = ax n. Then logy = log(ax n). So logy = loga + logx n (law1). Or logy = nlogx + loga (law3) . Now let Y = logy, X = logx, m = n, c = loga. This gives us Y = mX + c. Linear form.
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Linear Logarithmic Equations & y = axn Exponential form Suppose that y = axn Then logy = log(axn) So logy = loga + logxn (law1) Or logy = nlogx + loga(law3) Now let Y = logy, X = logx, m = n, c = loga This gives us Y = mX + c Linear form The linear form is easier to manipulate and can be converted back to the exponential form when required.
Ex1 lny NB: straight line with gradient 5 and intercept 0.69 m = 5 Using Y = mX + c 0.69 We get lny = 5lnx + 0.69 or lny = 5lnx + lne0.69 lnx or lny = 5lnx + ln2 or lny = lnx5 + ln2 Express y in terms of x. law3 or lny = ln2x5 law1 or y = 2x5
Ex2 log10y Using Y = mX + c 0.3 We get log10y = -0.3log10x + 0.3 log10y = -0.3log10x + log10100.3 log10x 1 log10y = -0.3log10x + log102 Find the formula connecting x and y. log10y = log10x-0.3 + log102 law3 log10y = log102x-0.3 law1 y = 2x-0.3 NB: straight line with intercept 0.3 Gradient = -0.3/1 = -0.3