Question

Can Somebody please convert this equation to a logarithmic equation?
42.07e^(0.75x) = y

Edit: "e" as in the constant e equal to 2.718218 and so on.

Answer 1

i think may be ,,e" the base of ligarithm naturale - yes ?

Answer 2

I edited the question jhonyy9.

Answer 3

where or how ? i dont see there nothing modified ?

Answer 4

first of all divide both sides by 42,07 so what will get ?

Answer 5

"e" as in the constant e equal to 2.718218 and so on.
y/42.07 = e^-0.75x
Does this work or do I need to multiply 42.07 by the constant e first?

Answer 6

Then would it be Ln(y/42.07) = -0.75x?
To convert that into a function that can be graphed using technology you use the change-base formula to make it Log(y/42.07)/Log(e)? I'm lost here.

Answer 7

but firtsly you wrote e^(0,75x)
how get this minus ?

Answer 8

My bad, that was a typo. The original equation is y = 42.07e^(-0.75x).

Answer 9

but you need to solve it for x ?

Answer 10

I just need to make it a logarithmic equation.

Answer 11

y = 42.07e^(-0.75x)

take at the ln of both sides:
ln y = ln(42.07) - 0.75 x

we could write that as
0.75 x = - ln y + ln 42.07
or
x= 4/3 ln(y) + 4/3 ln(42.07)

Answer 12

Why does one take the ln of both sides?

Answer 13

to keep it an equation, you have to "do the same thing" to both sides

Answer 14

Does the ln get rid of the constant e? Why did the exponent become a term by itself?

Answer 15

let me fix the typo:

or
x= -(4/3) ln(y) + (4/3) ln(42.07)

Answer 16

by definition
ln(e^x) = x

also, there is this property
log(ab) = log(a) + log(b)
(for any base, which includes ln i.e. base e)

so ln(4 e^x) = ln(4) + ln(e^x) = ln(4) + x

Answer 17

What level of math did you learn this in because my Algebra 2 teacher failed to mention the "ln(e^x) = x" definition.

Answer 18

Did they mention
\[ \log_b b^x =x \] ?

Answer 19

Not specifically but I do know that the exponent on be can become a coefficient and then l\[ \log_{b} b = 1\] . In the end x=x

Answer 20

ok, so
\[ \log_b b^x = x \log_b b = x \cdot 1= x \]

ln means \( \log_e \) so we do the same thing for
\[ \ln e^x = x \log_e e = x \cdot 1 = x \]

Answer 21

but to me "log" means "grab the exponent" as long as we have the "correct" base

Answer 22

That is understandable. Thank you for the hard work.