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20e^3x=100
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and another problem is 2^5x=24
2
its not 2
yes it is
-1.6
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for the 2nd question
i need all work
For the first one... Start off by dividing both sides by 20 to get the 'e' by itself \[\large e^{3x} = 5\] Take the natural log of both sides \[\large \ln(e^{3x}) = \ln(5)\] \[\large 3x = \ln(5)\] \[\large x = \frac{\ln(5)}{3}\] Second problem just like that
can you solve second
just bc it doesnt have e its confusing
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