11.5 Exponential and Logarithmic Equations

Transcription

1 .5 Eponentil nd Logrithmic Equtions In this section we wnt to solve equtions tht contin eponentil nd rithmic functions. Let s strt by reviewing some properties. Properties of Eponentil nd Logrithmic Equtions Let,, nd y be rel numbers such tht,, y 0,. y if nd only if y. y if nd only if y., ln e. ln, e. Then the following re true. Recll tht the lst two properties re true becuse the eponentil nd rithmic functions re inverse functions. We hve used these properties in clss to solve simpler equtions. We wnt to do more complicted equtions now. Before we do so, here is the procedure tht we will use. Solving Eponentil nd Logrithmic Equtions. To solve n eponentil eqution, isolte the eponentil epression, then tke the rithm of both sides of the eqution (using suitble bse) nd solve for the vrible.. To solve rithmic eqution, isolte the rithmic epression, then eponentite both sides of the eqution (using suitble bse) nd solve for the vrible. It is importnt to note tht when we solve rithmic eqution, we must check our nswer. The reson for this is tht the rithm hs domin restricted to only positive rel numbers. Hence, we need to check for zeros nd negtives in the rithm. Let s proceed to our equtions. We will review few esier ones first. Emple Solve 5 e 0. This is n eponentil eqution. Thus we must isolte the eponentil epression. 5 e e 0 5 Now lets tke the rithm of both sides. The bse we should use is e, since we get e ln e 5 ln5 ln5 ln5 Now just solve for. So the solution is ln5. 0. ln e. Thus,

2 Emple Solve. This time we hve rithmic eqution. So we must isolte the rithmic epression nd eponentite both sides. We will use bse to eponentite becuse. Here is the work. So our solution is. But we need to check to mke sure tht this does not violte the domin of our eqution. So plugging into our solution is. Lets proceed to more chllenging questions. Emple 50 Solve 0. t we get, which is oky. Hence We re gin deling with n eponentil eqution. So we must isolte the eponentil prt. We use techniques tht we lerned before. Lets begin by multiplying both sides of the eqution by the LCD (i.e. t ). This gives 50 0 Now it is just like before. We cn isolte, nd tke the rithm of both sides. Let s use bse so tht we cn use our properties. t 50 t 0 t 5 t 5 5 t 5 t

3 So, our solution is 5 t deciml equivlent nswer. Tht is,. But we must use the chnge of bse formul to get the 5 ln 5 t 0. ln Emple Solve. This is rithmic eqution. So we must isolte the rithm. However, there re two of the rithms there. But they re seprted by plus sign nd we know tht we cn condense rithms tht re seprted by plus or minus. We solve s follows 6 0 But checking in the originl eqution we get which violtes the domin, but checks out oky. Thus the solution is. 0 Notice in the lst emple we cn not just eponentite ech piece of n eqution. We cn only eponentite n entire side of the eqution., which is incorrect. The common mistke is Which gives Emple 5 or. Solve Agin we must isolte the rithm first. Creful to not eponentite ech piece of the eqution, but only eponentite the entire side.

4 So the solution is. Upon checking in the originl eqution, the solution is oky..5 Eercises Solve the following. Give both ect nd deciml pproimte nswers ln. ln. e ln 5 5. e ln ln , t.0 5 e t ln t e t ln e t. ln 55 e t 5e ln ln ln e t 000. ln

5 t e e ln ln e 5 ln ln e 5. ln ln (Hint: Use substitution) 5. e 0 (Hint: Use substitution) ln ln Find the - nd y-intercepts of the following. y y ln y e 0.. y 5. y Find the inverse of the following functions. y 6 e y. y 5. f 5. f 6. f. f. f e. f 5 0. f 6. f ln 5 Solve for the indicted vrible I M for I 0.. I 0 kt y Ce for k. 6. T kt Ts D0 e for t.. y c d be for d.. rt A Pe for t.. rt A Pe for r. kt y Ce for C.. H T kt Ts D0 e for k. 0. b c ph for H + y c d be for. y e for c.. y b h k for.

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