Home > Precalculus > Logarithmic Models

How do you solve #e^(x+6) = 4#?

Answer 1

Madelyn Anderson

Take the natural logarithm of both sides, then subtract #6# from both sides to find:

#x = log_e(4)-6 ~= -4.6137#

Taking #log_e# of both sides we get:

#log_e(4) = log_e(e^(x+6)) = x + 6#

Subtract #6# from both ends to get:

#x = log_e(4) - 6 ~= 1.3863 - 6 = -4.6137#

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Serenity Jones

To solve ( e^{x+6} = 4 ), you can take the natural logarithm (ln) of both sides of the equation to isolate ( x ):

[ \ln(e^{x+6}) = \ln(4) ]

Using the property of logarithms that ( \ln(e^a) = a ), where ( a ) is any real number:

[ x + 6 = \ln(4) ]

Then, solve for ( x ) by subtracting 6 from both sides:

[ x = \ln(4) - 6 ]

Therefore, the solution to the equation ( e^{x+6} = 4 ) is ( x = \ln(4) - 6 ).

By signing up, you agree to our Terms of Service and Privacy Policy

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.