1. ## [SOLVED] logrithimic question

it's

$\displaystyle 5^{x+1}=2(3^{2x})$

I have used base 10
and Im at x=log_5-log_2 over 2log_3-log_5/2

I expanded the bottom and cancelled -log_2

so I have log_5 over 2log3-log_5

I cant cancell the log base 5 because you cant make them have the same sign (neg. pos.) correct?

thank you.

2. Originally Posted by brentwoodbc
it's

$\displaystyle 5^{x+1}=2(3^{2x})$

I have to convert it to log form, then I think I have to use the formula
log_a over log_b
but I dont know how to convert something to logs with something on two sides of the = sign,

thank you.
Multiply out that two to give 6^{2x}

Taking logs of both sides (I will choose base e) and using the rule

$\displaystyle k ln(x) = ln(x^k)$

$\displaystyle (x+1)ln(5) = 2x ln(6)$ (using base 10 would give you $\displaystyle (x+1)log_{10}5 = 2xlog_{10}6$ and you can solve it in the same way)

Can you solve it from there? I don't see why you'd use the change of base rule for this problem

Spoiler:
$\displaystyle 2xln(6) = xln(5) + ln(5)$

$\displaystyle 2xln(6)-xln(5) = ln(5)$

$\displaystyle x(2ln(6)-ln(5)) = ln(5)$

$\displaystyle x = \frac{ln(5)}{2ln(6)-ln(5)}$

3. Originally Posted by e^(i*pi)
Multiply out that two to give 6^{2x}

Taking logs of both sides (I will choose base e) and using the rule

$\displaystyle k ln(x) = ln(x^k)$

$\displaystyle (x+1)ln(5) = 2x ln(6)$ (using base 10 would give you $\displaystyle (x+1)log_{10}5 = 2xlog_{10}6$ and you can solve it in the same way)

Can you solve it from there? I don't see why you'd use the change of base rule for this problem
I thought you couldnt multiply something out if it has a differnt power?

ie 2x2^2 = 8 not 16

4. Originally Posted by brentwoodbc
I thought you couldnt multiply something out if it has a differnt power?

ie 2x2^2 = 8 not 16
You can multiply the base but not the exponent. In the example you just put the base is 2 which would give 8. It is the same as 2^2 + 2^2 = 4 + 4 = 8.

If I were to do 3(2^2) which is obviously 12 it would be 2^2+2^2+2^2 = 4+4+4 = 12.

If it were a(b^n) then I'd get ab^n = b^n + b^n ... + b^n up to n=a

5. hm. my answer comes out close to yours. But the answer they give is

x= log2-log5 over log5-2log3

6. Originally Posted by brentwoodbc
hm. my answer comes out close to yours. But the answer they give is

x= log2-log5 over log5-2log3
$\displaystyle 5^{x+1}=2(3^{2x})$

taking logs of both sides

(x+1)log(5) = log(2(3^2x))

we can simplify the right hand side by using the law $\displaystyle log{(ab)} = log(a) + log(b)$ with a = 2 and b = 3^2x. Also expanding the left

$\displaystyle xlog(5) + log(5) = log(2) + 2xlog(3)$

combine x terms:

$\displaystyle -2xlog(3) + xlog(5) = log(2)-log(5)$

Factor

$\displaystyle x(-2log(3) + log(5)) = log(2)-log(5)$

and so $\displaystyle x = \frac{log(2)-log(5)}{log(5)-2log(3)}$

7. Thank you, I appreciate it

I did it a (slightly different way) and got

log5-log2 over 2log3-log5

is that the same thing? I know sometimes with fractions you can swap positions and negative signs and have the same value in a different order.

your way makes sense though so thanks agian.

8. Originally Posted by brentwoodbc
Thank you, I appreciate it

I did it a (slightly different way) and got

log5-log2 over 2log3-log5

is that the same thing? I know sometimes with fractions you can swap positions and negative signs and have the same value in a different order.

your way makes sense though so thanks agian.
Yeah, that's the same answer. It's the same as taking the negative of top and bottom which would cancel.