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# Urgent help needed with P2 logs! watch

1. How do you solve the following:

5^x+3 = 3^x-2 ?

The answer is -13.8, but I can't get it. Please can someone show me the proper method of solving these kind of equations asap so I can stop panicking?!

Thanks!
2. (Original post by Jen05)
5^x+3 = 3^x-2
5^(x + 3) = 3^(x - 2)
-> ln[5^(x + 3)] = ln[3^(x - 2)]
-> (x + 3)(ln5) = (x - 2)(ln3)
-> x(ln5) + 3(ln5) = x(ln3) - 2(ln3)
-> x(ln5) - x(ln3) = -2(ln3) - 3(ln5)
-> x(ln5 - ln3) = -[2ln3 + 3ln5]
-> x = -[2ln3 + 3ln5]/(ln5 - ln3)
-> x = -13.8 (3.S.F)
3. (Original post by Jen05)
How do you solve the following:

5^x+3 = 3^x-2 ?

The answer is -13.8, but I can't get it. Please can someone show me the proper method of solving these kind of equations asap so I can stop panicking?!

Thanks!
4. Take logs of both sides

gives,

(x+3)ln5 = (x-2)ln3
xln5 + 3ln5 = xln3 - 2ln3
x(ln5 - ln3) = -ln(5^3) - ln(3^2) = -ln125 - ln9 = -ln(125*9) = -ln(1125)
xln(5/3) = -ln(1125)
xln(3/5) = ln(1125)
xln(0.6) = ln(1125)
x = ln(1125)/ln(0.6) = -13.75
===============
5. lotsa replies
6. Thanks so so much guys! I understand now. For some reason I never realised you could change (x+3)ln5 into xln5 +3ln5. Hopefully that will come in handy in the exam!

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