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How do you solve #ln(e^(7x)) = 15 #?

Answer 1

Sophie Adams

#x=15/7#

This is for natural logarithms.

#ln(e^(7x))=15#

#ln(a^b)=bln(a)#

Then:

#ln(e^(7x))=15#

#7xln(e)=15#

#ln(e)=1# ( The logarithm of the base is always 1 )

#7x=15=>x=15/7#

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Answer 2

Caleb Alexander

To solve ( \ln(e^{7x}) = 15 ), we can use properties of logarithms and exponentials:

Recognize that ( \ln(e^{7x}) ) simplifies to ( 7x ), as the natural logarithm and the exponential function are inverses of each other.
So, the equation becomes ( 7x = 15 ).
Solve for ( x ) by dividing both sides by 7:

[ x = \frac{15}{7} ]

Therefore, the solution to ( \ln(e^{7x}) = 15 ) is ( x = \frac{15}{7} ).

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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.