Solving Algebra Problem: (7/4)^x = 5^(1-x)

[macbowes]

Alright, I just did my Math midterm and I came across this question.

(7/4)^x = 5^(1-x)

I would use Latex, but I find it confusing to use :(.

Anyways, what I ended up doing was using my graphing calculator and putting (7/4)^x in Y1 and 5^(1-x) in Y2. Then I just found where they intersect and obviously that value was X. The problem I'm having was the question stated to do it with algebra, but I don't know how. If someone could be so kind and show me, that would be swell.

EDIT: I put it in this section and not the homework section because it's not really homework; I've already done it, I just want to know how to do it via algebra.

EDIT2: Got it, thanks everyone!

 

[zgozvrm]

Try taking the LOG of both sides...

 

[Bill Simpson]

Hint: I expect you realize that +, -, * and / aren't going to do anything to powers and exponents. So... what other function have you been introduced to that does do something with powers and exponents?

I strongly urge you to not read what someone else writes just giving you the answer. Take my hint and think hard about this until you come up with one or more ideas. Then try those ideas, discard the ones that don't work, and keep at this until you discover the answer for yourself. The hint should be enough.

After you have found the answer then check it to see that it matches what your calculator gave. And after that think carefully about why you were not able to find the answer yourself during the test and what you can do to avoid that in the future.

 

[berkeman]

Alright, I just did my Math midterm and I came across this question.

(7/4)^x = 5^(1-x)

I would use Latex, but I find it confusing to use :(.

Anyways, what I ended up doing was using my graphing calculator and putting (7/4)^x in Y1 and 5^(1-x) in Y2. Then I just found where they intersect and obviously that value was X. The problem I'm having was the question stated to do it with algebra, but I don't know how. If someone could be so kind and show me, that would be swell.

EDIT: I put it in this section and not the homework section because it's not really homework; I've already done it, I just want to know how to do it via algebra.

Welcome to the PF.

It's still schoolwork, so it goes in the Homework Help forums. I've moved it for you.

 

[macbowes]

I suppose I should have posted what I've already tried to do (I worked on this question for over half an hour on my midterm).

Here's what I've tried so far:

(7/4)^x = 5^1-x

log₅(7/4)^x = 1-x

xlog₅(7/4) = 1-x

I'm not sure if I can bring the ^x to the front of the log, but that's what I did.

log₅(7/4) = (1-x)/x

Once again, I'm not sure if you're allowed to do this, but I divided the other side by x.

I then used change of base formula:

log(7/4)/log5 = (1-x)/x

0.3477 = (1-x)/x

0.3477x = 1-x

0.3477x + x = 1

1.3477x = 1

x = 1/1.3477

x = 0.801

That's what I tried to do, and I know I've made a mistake somewhere in there because that's not the answer I get when I do it via graphing.

EDIT:

Welcome to the PF.

It's still schoolwork, so it goes in the Homework Help forums. I've moved it for you.

Thanks :D.

EDIT2: Well, it turned out that I was doing right the whole time and I was getting the wrong answer due to me entering it wrong on my calculator.

 

[zgozvrm]

I would use Latex, but I find it confusing to use :(.

You could do it a couple of ways...

(7/4)^x = 5^{1-x}

\left( \frac{7}{4} \right)^x = 5^{1-x}


Hit the "Quote" button on my reply to see the LaTex code...

 

[zgozvrm]

x = 1/1.3477

x = 0.801

Looks fine, up until the final step. Try dividing that again...

(I think you mis-keyed a 2 instead of a 3).

 

[macbowes]

Looks fine, up until the final step. Try dividing that again...

Wow, haha. Of course it would come down to the easiest part.

Thanks :)

 

[zgozvrm]

Also, you can avoid the "change of base" formula by sticking with either Log_{10}[/tex] or Log_e[/tex] (I like to use Log_{10}[/tex] because it doesn't require the 2nd function key on my calculator) like this...<br /> <br /> <br /> (7/4)^x = 5^{1-x}<br /> <br /> Log \left((7/4)^x \right) = Log \left( 5^{1-x} \right)<br /> <br /> x Log(7/4) = (1-x) Log(5)<br /> <br /> \frac{1-x}{x} =\frac{Log(7/4)}{Log(5)}<br /> <br /> 1/x - 1 = \frac{Log(7/4)}{Log(5)}<br /> <br /> 1/x = 1 + \frac{Log(7/4)}{Log(5)}<br /> <br /> 1/x = 1 + 0.3477 = 1.3477<br /> <br /> x = 1/1.3477

 

[macbowes]

Also, you can avoid the "change of base" formula by sticking with either Log_{10}[/tex] or Log_e[/tex] (I like to use Log_{10}[/tex] because it doesn't require the 2nd function key on my calculator) like this...<br /> <br /> <br /> (7/4)^x = 5^{1-x}<br /> <br /> Log \left((7/4)^x \right) = Log \left( 5^{1-x} \right)<br /> <br /> x Log(7/4) = (1-x) Log(5)<br /> <br /> \frac{1-x}{x} =\frac{Log(7/4)}{Log(5)}<br /> <br /> 1/x - 1 = \frac{Log(7/4)}{Log(5)}<br /> <br /> 1/x = 1 + \frac{Log(7/4)}{Log(5)}<br /> <br /> 1/x = 1 + 0.3477 = 1.3477<br /> <br /> x = 1/1.3477

<br /> <br /> Oh, that's much better! I totally forgot that I could just make it base 10. I was stuck in the thinking that when going from exponential form to logarithmic I had to do this: a<sup>b</sup> = c -> log<sub>a</sub>c = b.

 

[zgozvrm]

And, you can combine the terms a bit more if you like:1/x = 1 + \frac{Log(7/4)}{Log(5)}

1/x = \frac{Log(5)}{Log(5)} + \frac{Log(7/4)}{Log(5)}

1/x = \frac{Log(5) + Log(7/4)}{Log(5)}

x = \frac{Log(5)}{Log(5) + Log(7/4)}

x = \frac{Log(5)}{Log(5) +\left( Log(7) - Log(4) \right)}

x = \frac{Log(5)}\left( Log(5) + Log(7) \right) - Log(4)}

x = \frac{Log(5)}{Log(35) - Log(4)}