Some Confusion with an Exponential Equation

[M83]

I'm reviewing for my final exam so one of the practice problems is:

5= 3^(x+5)

Here's my attempt at it:

ln 5= x+5 ln 3

ln 5 / ln 3 = x+5
(ln 5 / ln 3)-5= x
1.46-5 ≈ x
-3.54 ≈ x

I checked my answer and I get 3^1.46 ≈ 4.97 so rounding it up gives me 5 since I rounded off 1.46. However when I plug this equation into Mathway it gives me an answer of -3.39. I tried figuring out why (I know if you pay a fee you can view their steps, but I don't have the money at the moment) and eventually came to the conclusion that this is how they did it:

5= 3^(x+5)

x+5 = ln 5
x = ln 5 - 5
x ≈ -3.39

But when I check their answer I get 3^1.61 ≈ 5.86

I think Mathway is wrong, but maybe I'm missing something. Which method is the correct one for solving this equation? Thanks for any help I receive.

 

[Bohrok]

Don't worry, your answer is correct.
Also, give Wolfram|Alpha a try; it's free and it shows steps (although they're not always very helpful depending on the problem).

 

[M83]

Thanks Bohrok. I'll give Wolfram a try.

 

[Mark44]

I'm reviewing for my final exam so one of the practice problems is:

5= 3^(x+5)

Here's my attempt at it:

ln 5= x+5 ln 3

The equation above needs parentheses.

What you meant was
ln 5= (x+5) ln 3

Because of the higher precedence of multiplication over addition, what you wrote would be interpreted as
ln 5= x+ (5 ln 3)

ln 5 / ln 3 = x+5
(ln 5 / ln 3)-5= x
1.46-5 ≈ x
-3.54 ≈ x

I checked my answer and I get 3^1.46 ≈ 4.97 so rounding it up gives me 5 since I rounded off 1.46. However when I plug this equation into Mathway it gives me an answer of -3.39. I tried figuring out why (I know if you pay a fee you can view their steps, but I don't have the money at the moment) and eventually came to the conclusion that this is how they did it:

5= 3^(x+5)

x+5 = ln 5
x = ln 5 - 5
x ≈ -3.39

But when I check their answer I get 3^1.61 ≈ 5.86

I think Mathway is wrong, but maybe I'm missing something. Which method is the correct one for solving this equation? Thanks for any help I receive.

 

[M83]

The equation above needs parentheses.

What you meant was
ln 5= (x+5) ln 3

Because of the higher precedence of multiplication over addition, what you wrote would be interpreted as
ln 5= x+ (5 ln 3)

Thanks for the correction.

 

[SteveL27]

I'm reviewing for my final exam so one of the practice problems is:

5= 3^(x+5)

5 = 3^x * 3^5

5/(3^5) = 3^x

x = log_3(5/(3^5)) whatever that turns out to be. Let's go further ...

x = log_3(5/(3^5)) = log_3(5) - log_3(3^5)

= log_3(5) - 5

= -3.53502647928207283280295959232135960369206763333395...

that last step from Wolfram Alpha. So I think what you had initially is correct.

 

[HallsofIvy]

You could also take the logarithm directly:
[tex]5= 3^{x+ 5}[/tex]
[tex]log(5)= log(3^{x+5})= (x+ 5)log(3)[/tex]
so that
[tex]x+ 5= \frac{log(5)}{log(3)}[/tex]
and then
[tex]x= \frac{log(5)}{log(3)}- 5[/tex].

Are you required to write a decimal answer (which can only be approximate)? I would leave the answer as the above fraction.

 

[stef6987]

your answer is correct, with these questions always take the ln/e and solve from their, quite straight forward :)