help!

[thetomps]

If 2^3x = 5^x + 1, then correct to two decimal places, x =

-12.11
-6.39
-2.71
-0.76
1.71
3.42
5.03

 

[stapel]

You have posted:

. . . . .2³x = 5^x + 1

Is this what you meant, or is the equation more along the lines of:

. . . . .2³x = 5^{x + 1}

...or:

. . . . .2^3x = 5^x + 1

...or:

. . . . .2^3x = 5^{x + 1}

...or something else?

When you reply with clarification, please include all the steps you have tried thus far. Thank you.

Eliz.

 

[thetomps]

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

If i knew how to start it or do it I wouldn't have asked for help.

 

[stapel]

I'm going to guess the equation is:

. . . . .2^3x = 5^{x + 1}

...since leaving the x out of the exponent on the left-hand side would be rather unusual (though not impossible) for this sort of exercise.

To get the variable "x" out of the exponents in this exponential equation, you need to start by taking logs of either side. Are you familiar with logarithms?

Thank you.

Eliz.

 

[thetomps]

not really........

 

[stapel]

Ouch. :shock:

Without logs, I don't see any way of solving this. They shouldn't have asked, if they haven't covered this in class yet. Unless....

When they say "correct to two decimal places", are you maybe supposed to graph the two sides on your graphing calculator, as:

. . . . .Y1 = 2^(3X)
. . . . .Y2 = 5^(X+1)

...and then "ZOOM" in or use "TRACE" or "ISECT" or something, to find the intersection accurate to two places? (Actually, you wouldn't even have to get very accurate on the calculator, since all you're really doing is seeing which listed option matches the graph best.)

Just a thought....

Eliz.

 

[thetomps]

-0.76

 

[stapel]

To check the answer to any "solving" exercise, plug it back into the original problem. So:

. . . . .2^3(-0.76) = 2^-2.28 = 0.2058977...

. . . . .5^{(-0.76) + 1} = 5^0.24 = 1.471474...

Since the two sides of the equation do not evaluate to the same number, "x = -0.76" is unlikely to be correct.

Eliz.

 

[thetomps]

3.42

 

[pka]

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

\(\displaystyle x\ln (8) = (x + 1)\ln (5)\)

\(\displaystyle x = \frac{{\ln (5)}}
{{\ln (8/5)}}\)

 

[thetomps]

pka, I don't understand?

 

[pka]

Maybe this problem is just over your head.
There is no way to solve without knowledge of logarithms!
Whoever or whatever gave you this problem expected you to know logarithms.
Seeing that you do not know them, I gave you the answer.
You can use a common calculator to evaluate the answer.

 

[stapel]

pka used logarithms, but since your class hasn't covered that, just use the graphical method instead. (I can't think of any way, other than graphing, that they could expect you to use before having covered logs.)

By the way, yes, your latest answer is correct.

Eliz.

 

[pka]

Eliz, I have a very hard time thinking that this problem would be given in any class that had not studied logarithms. If it were, then that is just irresponsible! Graphing is a great problem solving technique. But in this case I cannot see it being used.

 

[stapel]

Eliz, I have a very hard time thinking that this problem would be given in any class that had not studied logarithms. If it were, then that is just irresponsible! Graphing is a great problem solving technique. But in this case I cannot see it being used.

You're right, of course, that "by graphing" isn't the best way to do this, and it certainly doesn't teach the student much. But if the student's class hasn't covered logs yet, I see no alternative.

I share your doubts about logs not yet having been covered, but what alternative could we suggest -- that wouldn't just generate more hostility? Students get mad enough when you ask them please to show any work of their own, or to post the answer they're claiming to have found that they want you to check. How well do they take the suggestion that maybe they should have been paying a little more attention for the last month or so of class?

And, to be fair, "cutting edge" "reform" educationist philosophy does elevate graphing (with calculators, of course; never by hand) far above the actual algebra. This poster could, unfortunately, be in that sort of a classroom, in which case, "by graphing" is exactly what they're expecting.

Eliz.

 

[pka]

That is fair enough!
But I can tell you that I have been an active member or the “calculus reform movement” since 1985. I have never seen graphing used in the context.
If this is a case where it is, then it is overkill!

 

[stapel]

I have been an active member or the “calculus reform movement” since 1985. I have never seen graphing used in the context.

I wasn't studying for an education degree until about ten years later, so I may have encountered a later variant of the school of thought. And of course, what you've been a part of "up" in calculus might not have filtered "down" into algebra in the intended form. Hard to say.

If this is a case where it is, then it is overkill!

Can't argue with ya there!

Eliz.