Can anyone help me with logarithms that contain the number"e(2.178 thing)"?

Question

I'm not sure if I'm doing it right... I have diff. ones...
1.2e^(-5x)+2.6=3
1.2e^(-5x)=.4
e^(-5x)=.333333333333333333
ln .3333 = -5x
x= .2199

Would this be right????

and I need help with... -16 + 0.2(10)^x = 35... I'm not really sure what to do here.

Answer 1

first one is right

second one use base-10 logarithms
-16 + 0.2(10)^x = 35
0.2 * 10^x = 51
10^x = 255
x = log(10) 255 = 2.4065

Answer 2

There is no need to write "0.3333..." You should use fractions, since any decimals like this that you write are always going to have some round-off error. Fractions are always exact. Dividing both sides by 1.2 and reducing the fraction, you get
e^(-5x) =0.4 / 1.2
e^(-5x) = 4 / 12
e^(-5x) = 1/3

Now apply the definition of logs, more specifically here the natural log:
Ln (1/3) = -5x
x = -[ Ln(1/3) ] / 5

And NOW, if you wanted to put the answer in decimal form (and I don't know why you would unless they asked), you can use the calculator to figure that out.

Answer 3

It looks pretty reasonable. Just don't get carried away by the division of 0.4/1.2, the answer is 1/3.

If you answered the first one, the second shouldn't be all that difficult.
Step 1, clear non-powers to one side
0.2 10^x = 51
Divide by 0.2 10^x = 205
Find the log of both sides (base 10)
x = 2.31 approximately

BTW, the exponent "e" is a very important number in mathematics, so get it right. It is about 2.718.
Another sneaky hint e^2.3 is about 10