How do you solve #10 = e ^ x#?

Answer 1

I found: #x=2.3025#

One way is to take the natural log of both sides (which can be evaluated using a pocket calculator) to get: #ln10=lne^x# where from the definition of log: #lne^x=x# so: #x=ln10=2.3025#

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Answer 2

Elijah Allen

#x = ln(10)#

To get rid of the base #e#, we can take the natural log of both sides:

#10 = e^x -> ln(10) = ln(e^x) -> x = ln(10)#

This value is irrational, so the only exact way to write it is #ln(10)#.

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Answer 3

Elias Ackerman

Take the natural log of both sides of the equation ...

#ln10=ln(e^x)#

Now, simplify and solve for x ...

#ln10=x#

#x~~2.3026#

Verify ...

#e^2.3026=10.000#

hope that helped

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Answer 4

Caleb Alexander

To solve ( 10 = e^x ), take the natural logarithm (ln) of both sides to isolate ( x ):

[ \ln(10) = \ln(e^x) ]

Since ( \ln(e^x) = x ), we have:

[ x = \ln(10) ]

So, ( x = \ln(10) ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.