solve for an exponent - logarithm: 3000(1+0.035)^x=5026

[idontknowcookie]

Hello,

for an assignment I made an equation, now I have to solve it for x.

I can't remember how I did it in high school, I just now it's something with logarithm but I don't know how to do it exactly. I never got what logarithms are though.

3000(1+0.035)^x=5026

Thank you for your explanation!

 

[Deleted member 4993]

Hello,

for an assignment I made an equation, now I have to solve it for x.

I can't remember how I did it in high school, I just now it's something with logarithm but I don't know how to do it exactly. I never got what logarithms are though.

3000(1+0.035)^x=5026

Thank you for your explanation!

3000(1+0.035)^x=5026

(1+0.035)^x=5026/3000

ln[(1+0.035)^x] = ln[5026/3000]

x * ln[(1+0.035)] = ln[5026/3000]

Continue....

 

[idontknowcookie]

3000(1+0.035)^x=5026

(1+0.035)^x=5026/3000

ln[(1+0.035)^x] = ln[5026/3000]

x * ln[(1+0.035)] = ln[5026/3000]

Continue....

Thank you for the help.

So, the last step is ln[5026/3000]/ln[1+0.035]=x=15.0

A further question: Why did you choose ln and not log? What would have changed if you put log there?

 

[stapel]

Why did you choose ln and not log? What would have changed if you put log there?

Science-y and math-y people tend to use the natural log, because it keeps cropping up in so many real-world contexts. Either log (base-e or base-10, or any other base, for that matter) would have been fine, and would have led to the same result, by the change-of-base formula:

. . . . .\(\displaystyle \ln(a)\, =\, \dfrac{\log(a)}{\log(e)}\)

. . . . .\(\displaystyle \dfrac{\ln(a)}{\ln(b)}\, =\, \dfrac{\left(\, \dfrac{\log(a)}{\log(e)} \,\right)}{\left(\, \dfrac{\log(b)}{\log(e)} \,\right)}\, =\, \left(\dfrac{\log(a)}{\log(e)}\right)\,\)\(\displaystyle \left( \dfrac{\log(e)}{\log(b)} \right)\, =\, \dfrac{\log(a)}{\log(b)}\)