Quick question regarding a problem with exponentials

[imdone]

Hi, I just completely forgot how to do these type of questions since I haven't been into these maths for a decade, literally.

This type of exponential problems which I'm truly disappointed in my brain.
(a^x+n)+(b^x+m)=(c^x+i)+(d^x+j)
I believe there is a very easy way to solve this but it's just on the tip of my brain.

Example: 6.3^x+2 - 4.5^x+3 = 3^x+4 - 5^x+4; find x
Could anyone here tell me the theory or a way that can effectively solve these problems.
Thanks

 

[fresh_42]

Hi, I just completely forgot how to do these type of questions since I haven't been into these maths for a decade, literally.

This type of exponential problems which I'm truly disappointed in my brain.
(a^x+n)+(b^x+m)=(c^x+i)+(d^x+j)
I believe there is a very easy way to solve this but it's just on the tip of my brain.

Example: 6.3^x+2 - 4.5^x+3 = 3^x+4 - 5^x+4; find x
Could anyone here tell me the theory or a way that can effectively solve these problems.
Thanks

There is no way to solve this type of equation other than a numerical, i.e. an algorithm that computes a number that comes as close as you want, but still not exactly, e.g. by nested intervals.

 

[Mark44]

Example: 6.3^x+2 - 4.5^x+3 = 3^x+4 - 5^x+4; find x

Just to be clear, what you wrote is this:
$6.3^x + 2 - 4.5^x + 3 = 3^x + 4 - 5^x + 4$

You didn't happen to mean this, did you?
$6.3^{x + 2} - 4.5^{x + 3} = 3^{x + 4} - 5^{x + 4}$

 

[imdone]

There is no way to solve this type of equation other than a numerical, i.e. an algorithm that computes a number that comes as close as you want, but still not exactly, e.g. by nested intervals.

hmm, I'm fairly certain I used to solve these in high school with a different approach. By that, I mean minimise the equation into 'log term' and use a calculator, but not using it at the start.

Just to be clear, what you wrote is this:
$6.3^x + 2 - 4.5^x + 3 = 3^x + 4 - 5^x + 4$

You didn't happen to mean this, did you?
$6.3^{x + 2} - 4.5^{x + 3} = 3^{x + 4} - 5^{x + 4}$

Yes, sorry for the inconvenience.

 

[Mark44]

Just to be clear, what you wrote is this:
$6.3^x + 2 - 4.5^x + 3 = 3^x + 4 - 5^x + 4$

You didn't happen to mean this, did you?
$6.3^{x + 2} - 4.5^{x + 3} = 3^{x + 4} - 5^{x + 4}$

Yes, sorry for the inconvenience.

OK, also from what you wrote I'm assuming that on the left side 6.3 is being raised to the power x + 2 and 4.5 is being raised to the power x + 3. Is my interpretation accurate, or are you using . to indicate multiplication?

IOW, is this the problem you're asking about?
$6 \cdot 3^{x + 2} - 4 \cdot 5^{x + 3} = 3^{x + 4} - 5^{x + 4}$

I suspect that this is really the problem you meant to ask. If so, I get a solution of x = -3.

When you post a question involving mathematics, it's important to write it in a way that accurately communicates what you're working on. In this case, missing parentheses and using . for multiplication instead of * made it difficult to know what the problem was.

 

[Adirdagal]

Logarithms should do the trick. Given that log(a·b^(x+c))=log(a)+(x+c)·log(b) it seems to be a solvable problem.

 

[fresh_42]

Logarithms should do the trick. Given that log(a·b^(x+c))=log(a)+(x+c)·log(b) it seems to be a solvable problem.

No, because what is true for one term, fails for sums. There is no formula for $\log (a +b)$.

 

[Adirdagal]

No, because what is true for one term, fails for sums. There is no formula for $\log (a +b)$.

Quite right.

 

[imdone]

I suspect that this is really the problem you meant to ask. If so, I get a solution of x = -3.

Yes, sorry, but I'd love to know how you did that step by step or 'a quick easy way' to solve such problems?
As I mentioned 'I used to solve these in high school', this exact problem was given to me and I solved it fairly easily, but not now.

I hope you forgive me for the roughness of those messages, and I'm looking forward for your reply.

Thanks

 

[Mark44]

Assuming this is the problem you're asking about,
$6 \cdot 3^{x + 2} - 4 \cdot 5^{x + 3} = 3^{x + 4} - 5^{x + 4}$

$3^{x + 2} = 3^x \cdot 3^2 = 9 \cdot 3^x$, so $6 \cdot 3^{x + 2} = 6 \cdot 9 \cdot 3^x = 54 \cdot 3^x$
Carry out this same type of operation on all four terms in your equation, and then combine all of the $3^x$ terms on one side of the equation, and all of the $5^x$ terms on the other side. Give it a shot and we'll help you, but forum rules prevent me from just working the problem through.

 

[imdone]

$3^{x + 2} = 3^x \cdot 3^2 = 9 \cdot 3^x$, so $6 \cdot 3^{x + 2} = 6 \cdot 9 \cdot 3^x = 54 \cdot 3^x$

Sorry, it was 6.3x+2−4.5x+3=3x+4−5x+4 not 6⋅3x+2−4⋅5x+3=3x+4−5x+4
and also I'd like to know 'the easy way' or at least how to generally solve these problems not only 'this' problem.
Thanks, if you could help me.

 

[Mark44]

Sorry, it was 6.3x+2−4.5x+3=3x+4−5x+4 not 6⋅3x+2−4⋅5x+3=3x+4−5x+4
and also I'd like to know 'the easy way' or at least how to generally solve these problems not only 'this' problem.
Thanks, if you could help me.

Now we're going backwards. The thread title indicates that the problem involves exponents, but what you wrote above doesn't show that at all.

Just looking at the first part of the equation you wrote, 6.3x+2, you wrote this earlier as 6.3^x+2. If you mean this to be $6.3^{x + 2}$, it should be written using TeX as I did, or as 6.3^x+2, using the BBCode capabilities on this site, or as 6.3^(x + 2). Same for the other terms in your equation.

If this is the equation you're interested in solving, $6.3^{x + 2} - 4.5^{x + 3} = 3^{x + 4} - 5^{x + 4}$, I can't think of any easy way to solve it.

 

[imdone]

It is

$6.3^{x + 2} - 4.5^{x + 3} = 3^{x + 4} - 5^{x + 4}$

I assume you understand the question since my first post, however, let's put that aside I'll focus on the main topic.

So, basically you tell me you don't know the 'easy way' to solve it, but how would you solve it?
the easiest solution you could think of might be the solution I've been looking for

As always, thanks

 

[Mark44]

So, basically you tell me you don't know the 'easy way' to solve it, but how would you solve it?
the easiest solution you could think of might be the solution I've been looking for

As @fresh_42 said in post #2, the only way to solve this equation is by a numerical algorithm of some kind.