How would you go about solving: 6 = e^(1+b) + B 2 = e^b +B where e = euler's number, not an unknown Is it something to do with subtracting? I just keep getting myself into a dead end. Thanks
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Originally Posted by freswood How would you go about solving: 6 = e^(1+b) + B 2 = e^b +B where e = euler's number, not an unknown Is it something to do with subtracting? I just keep getting myself into a dead end. Thanks To solve: you rewrite them as: , which leaves you with a pair of linear simultaneous equations in and , which can be solved in the usual manner. RonL
How would you finish off what you've started?
Originally Posted by freswood How would you finish off what you've started? Subtract: , so: , and substituting back into the second equation: . Finally: RonL
Thanks so much! It kinda confused me because I wasn't expecting the answer to be that messy. Whole numbers all the way!
Hello, freswood! Solve: . This is a variation of Captain Black's solution . . . We have: . Divide the equations: . . . Then: . . . And: . . . Therefore: . Substitute into (2): . . . . . . Therefore: .
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