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2^x = e^5
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umm Y dont u just log both sides?
Seems like the simplest option to me
unless you need to use another method
algebraically you can use expansion..and solve it ...like the expansion of \[a ^{x}\] and \[e ^{x}\] but go with logs ..it would be easier
ln(2^x) = 5
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= x ln(2)
I'm getting x = 5/ln2. Is that right?
0.3010x=5 now solve for x
yes
@tylendel Yes
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Awesome. Thanks guys!
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