What are the steps to solve an equation with logarithms on both sides?

[kalistella]

Hi

I need some help!

Doing A'level logarithms and am stuck.

log3(x-1)=log8(x+1)

So far, I have done these steps...
(x-1)log3=(x+1)log8
xlog3-log3=xlog8+log8
xlog3-xlog8-log3-log8

Not sure what to do next to get the solution for x. Everything I do gets the wrong answer. I need an explanation in simple terms!

Any help greatly appreciated!

 

[HallsofIvy]

Hi

I need some help!

Doing A'level logarithms and am stuck.

log3(x-1)=log8(x+1)

Do you mean log(3(x-1)), log₃(x-1) or log 3^x-1?
(If you can't do subscripts or superscripts, use log_3(x-1) or log 3^(x-1).)

So far, I have done these steps...
(x-1)log3=(x+1)log8

Okay, it must be log(3^x-1)= log(8^x+1)

xlog3-log3=xlog8+log8
xlog3-xlog8-log3-log8

Whoops, what happened to the "="?? I assume you meant
xlog(3)- xlog(8)= log(8)- log(3)

Not sure what to do next to get the solution for x. Everything I do gets the wrong answer. I need an explanation in simple terms!

Any help greatly appreciated!

xlog(3)- xlog(8)= log(8)- log(3)
x(log(3)- log(8))= log(8)- log(3)
x= (log(8)- log(3))/(log(3)- log(8))

 

[Architeuthis Dux]

Hello! Logarithms can definitely be tricky, but don't worry, I'm here to help. Let's break down the steps you've already taken and see where you may have gone wrong.

First, you correctly used the property of logarithms that states log(a^b) = blog(a). This means that you can rewrite log3(x-1) as (x-1)log3 and log8(x+1) as (x+1)log8.

Next, you multiplied both sides by log3 and log8, which is also correct. This gives you the equation xlog3 - log3 = xlog8 + log8.

But here is where you may have made a mistake. When you have an equation with logarithms on both sides, you can't just subtract them like you would with regular numbers. Instead, you need to use the logarithm property log(a) + log(b) = log(ab).

So, let's apply this property to your equation. We can rewrite xlog3 - log3 as log3(x^x) and xlog8 + log8 as log8(x^x). This gives us the equation log3(x^x) = log8(x^x).

Now, we can use the property of logarithms again, log(a) = log(b) if and only if a = b. This means that since both sides of our equation have the same argument (x^x), we can set them equal to each other.

So, we have x^x = x^x. This may seem like we haven't made any progress, but it actually means that any value of x will satisfy this equation. So, the solution for x can be any real number.

I hope this explanation helps you understand logarithms a bit better. Keep practicing and don't get discouraged, you'll get the hang of it!