How to use common or natural logarithms to solve the exponential equation

[fawazalusail]

3e^(6x+4)=9
----------------

e^(6x+4)/3 = 9/3

ln e^(6x+4)=ln3

6x+4=ln3

6x+4-4=(ln3)-4

6x=(ln3)-4

6x/6=(ln3)-4/6

The answer is x=-0.4836.<<==== How!! when i divide by 6 I get this on my calculator -0.431945622. Please explain how we get that answer. Thanks

 

[mmm4444bot]

3e^(6x+4)=9
----------------

e^(6x+4)/3 = 9/3

ln e^(6x+4)=ln3

6x+4=ln3

6x+4-4=(ln3)-4

6x=(ln3)-4

6x/6=(ln3)-4/6

The answer is x=-0.4836.<<==== How!! when i divide by 6 I get this on my calculator -0.431945622. Please explain how we get that answer. Thanks

You are not following the correct Order of Operations.

(I cannot see what you're doing with unknown calculator). :cool:

First calculate ln(3) - 4 = -2.901387711

Now divide by 6

 

[lookagain]

6x/6=(ln3)-4/6

The answer is x=-0.4836.<<==== How!! when i divide by 6 I get this on my calculator -0.431945622. Please explain how we get that answer. Thanks

I know what you did. You either typed

(ln3) - 4/6 or

(ln(3)) - 4/6 or

ln(3) - 4/6


On a usual calculator that displays parentheses on a screen, type

"(ln(3) - 4)/6."


It can do the calculation all at once.

What you typed in equivalent to \(\displaystyle ln(3) \ -\ \dfrac{4}{6}.\)


But you mean to indicate \(\displaystyle \dfrac{ln(3) - 4}{6}.\)