# Solving Logarithms help

3 * 2^3x = 7 * 3^(3x-2)

i’ve tried to solve this multiple times and can’t get it right, the answers 1.11 and i keep getting 1.53, i’m not sure what i’m doing wrong
Original post by user987789
3 * 2^3x = 7 * 3^(3x-2)

i’ve tried to solve this multiple times and can’t get it right, the answers 1.11 and i keep getting 1.53, i’m not sure what i’m doing wrong

You need same basis to calculate powers with each other. Have you done anything for it?
i put all the terms to ln
i did
3 * 2^3x = 7 * 3^(3x-2)
ln3 * 3xln2 = ln7 * (3x-2) * ln3
ln3 * ln2 / ln7 * ln3 = 3x-2/3x
Original post by user987789
i did
3 * 2^3x = 7 * 3^(3x-2)
ln3 * 3xln2 = ln7 * (3x-2) * ln3
ln3 * ln2 / ln7 * ln3 = 3x-2/3x

you've got your log rules messed up! ln(AB) = ln(A) + ln(B) etc. I get 1.10977 approx when I bung my solution into Excel
Original post by user987789
i did
3 * 2^3x = 7 * 3^(3x-2)
ln3 * 3xln2 = ln7 * (3x-2) * ln3
ln3 * ln2 / ln7 * ln3 = 3x-2/3x

When you take the natural log on both sides, it should become:
ln(3 * 2^3x) = ln(7 * 3^(3x-2))

Which, because of the log rule ln(x * y) = ln(x) + ln(y), becomes:
ln(3) + ln(2^3x) = ln(7) + ln(3^(3x-2))

And then it looks like you know how to solve it from there. Let me know if you need more help
Original post by BlueBazooka
Original post by user987789
i did
3 * 2^3x = 7 * 3^(3x-2)
ln3 * 3xln2 = ln7 * (3x-2) * ln3
ln3 * ln2 / ln7 * ln3 = 3x-2/3x

When you take the natural log on both sides, it should become:
ln(3 * 2^3x) = ln(7 * 3^(3x-2))

Which, because of the log rule ln(x * y) = ln(x) + ln(y), becomes:
ln(3) + ln(2^3x) = ln(7) + ln(3^(3x-2))

And then it looks like you know how to solve it from there. Let me know if you need more help

i tried to correct it

3 * 2^3x = 7 * 3^(3x-2)
ln3 + 3xln2 = ln7 + (3x-2)ln3
3xln6 = (3x-2)ln21
3x-2/3x = ln6/ln21
x-2 = 0.5885190554
x= 2.5885190554
but i still don’t understand what i’m doing wrong
Original post by user987789
When you take the natural log on both sides, it should become:
ln(3 * 2^3x) = ln(7 * 3^(3x-2))

Which, because of the log rule ln(x * y) = ln(x) + ln(y), becomes:
ln(3) + ln(2^3x) = ln(7) + ln(3^(3x-2))

And then it looks like you know how to solve it from there. Let me know if you need more help

i tried to correct it

3 * 2^3x = 7 * 3^(3x-2)
ln3 + 3xln2 = ln7 + (3x-2)ln3
3xln6 = (3x-2)ln21
3x-2/3x = ln6/ln21
x-2 = 0.5885190554
x= 2.5885190554
but i still don’t understand what i’m doing wrong
you can't combine logs like that! Put all the 3x terms on one side and solve as a normal linear equation
Original post by user987789
When you take the natural log on both sides, it should become:
ln(3 * 2^3x) = ln(7 * 3^(3x-2))

Which, because of the log rule ln(x * y) = ln(x) + ln(y), becomes:
ln(3) + ln(2^3x) = ln(7) + ln(3^(3x-2))

And then it looks like you know how to solve it from there. Let me know if you need more help

i tried to correct it

3 * 2^3x = 7 * 3^(3x-2)
ln3 + 3xln2 = ln7 + (3x-2)ln3
3xln6 = (3x-2)ln21
3x-2/3x = ln6/ln21
x-2 = 0.5885190554
x= 2.5885190554
but i still don’t understand what i’m doing wrong

On the third line, unfortunately you can't combine logs when one or both of them have a coefficient.

The rule log(a) + log(b) = log(a * b) is true, and so is the reverse.
But in your case, you have something like log(a) + C*log(b), and that can't be combined into C*log(ab)

Instead, your third line of working should be something like:
3xln(2) - 3xln(3) = ln7 - ln3 - 2ln3

And then you can factorise out the x term and isolate it to solve. Hope this helps
(edited 5 months ago)