Monday I have an exam and I don’t fully understand logarithms.
I have trouble solving this equation:
10^(4x-1) = 5 . 2^x
What I have come up with:
10^(4x-1) = 5 . 2^x
log 10^(4x-1) = log 5 . 2^x
4x-1 . log 10 = log 5 . 2^x
4x-1 = log 5 . 2^x
4x-1 = 1/2 . 2^x
x-1 = (1/2 . 2^x)/4
(x-1)/(2^x) = (1/2)/4
(x-1)/(2^x) = 1/8
Is there someone who understands logarithms who can explain me where I went wrong and how to solve the equation?
I really appreciate any help!
Thanks,
Peter
Although this might not be the best place to ask homework problems…
when you take log on both sides, log( 5 . 2^x)=log 5 + log (2^x) = log 5 + x . log 2
careful about parenthesis.
So you can take
2 to the power of x (2^x)
and add x to log 5? Or is that the logarithm of 5 + x?
Damn, not only is my teacher a bulldog, but he’s also unable to teach. He does know his math, though.
The improtant thing is the following:
ln(x * y) = ln(x) + ln(y), where x and y are positive, real numbers.
As homebody said, here you have ln(52^x) = ln(5) + xln(2). So, the solution for x is (ln(5) + ln(10))/(4*ln(10) - ln(2)) or, using our rule you can also write it as ln(50) / ln(5000), which is probably more beautiful.
Oh, I just noticed that I used the ln symbol - don’t worry about that, though, it does not matter here.