algebra

[sammar]

can someone show me how to solve formulas with exponential like this one 100,000(1+x)^20=329,575

 

[Loren]

Example..

3(1+x)[sup:26o21uxy]5[/sup:26o21uxy]=18

(1+x)[sup:26o21uxy]5[/sup:26o21uxy] = 6

Take the log of both sides.

log(1+x)[sup:26o21uxy]5[/sup:26o21uxy] = log 6

5 log(1+x) = 0.778

log(1+x) = 0.778/5

log(1+x) = 0.1556

1+x = 10[sup:26o21uxy]0.1556[/sup:26o21uxy]

x = 10[sup:26o21uxy]0.1556[/sup:26o21uxy] - 1 = 0.429...

Check 3(1+x)[sup:26o21uxy]5[/sup:26o21uxy] = 3(1+0.43)[sup:26o21uxy]5[/sup:26o21uxy]= 3(1.43)[sup:26o21uxy]5[/sup:26o21uxy]= 3(5.958)=17.9... or 18 rounded.

Carry your decimals out further than two places for more accuracy.

 

[lookagain]

Example..

3(1+x)[sup:3joaobos]5[/sup:3joaobos]=18

(1+x)[sup:3joaobos]5[/sup:3joaobos] = 6

Or sammar, avoid taking logarithms after the last step shown in the quote box.

Take the fifth root of each side:

\(\displaystyle \sqrt[5]{(1 + x)^5} = \sqrt[5]{6}\)

\(\displaystyle 1 + x = \sqrt[5]{6}\)

\(\displaystyle x = \sqrt[5]{6} - 1\)

\(\displaystyle x \approx 0.4310\)

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Loren,

1) you rounded for \(\displaystyle log(6),\) and then you rounded for \(\displaystyle 10^{0.1556} - 1\).
This is an example of "the error of successive rounding."

2) \(\displaystyle 10^{0.1556} - 1 = 0.430869...\) instead of your value.

3) My value is even different from what I posted immediately above,
because I did not round until the end of the calculation.

 

[mmm4444bot]

"the error of successive rounding"

It's not really an error, if one qualifies their rounding with a specific statement, as Loren did.

But, I agree with your point. If one uses a decent calculator, one might as well carry all of the machine digits through to the end versus intermediary rounding. If one uses paper and pencil, one should decide the precision ahead of time and round intermediate results accordingly.

 

[lookagain]

Re:

"the error of successive rounding"

It's not really an error, if one qualifies their rounding with a specific statement, as Loren did.

That is the official name of it. You don't round until the end, which she did. Had she not
rounded until the end, then she would have gotten the correct three digits to the right of
the decimal point provided her calculator or computer was/is sufficient for the calculation.
Her statement about using two more digits doesn't guarantee a better answer, because her
own value of \(\displaystyle 0.429...\) is not correct beginning at the hundredths digit.
She lucked into it being rounded to the nearest hundredths as \(\displaystyle 0.43\)

And, in her check, she used the wrong value for \(\displaystyle 1.43^5\). It is \(\displaystyle 5.97971\) (correct
to the fifth digit to the right of the decimal point), so the line should have read as

Check \(\displaystyle 3(1 + x)^5 = 3(1 + 0.43)^5 = 3(1.43)^5 \approx 3(5.980) = 17.94\) or \(\displaystyle 18\) rounded.

 

[mmm4444bot]

Perhaps, I misunderstood Loren's intention about more digits more precision. Please excuse me.

 

[Loren]

The problem is with typing out a non-rounded decimal number. I erred by not being consistent in the level of my rounding. I was trying to show a procedure and got lazy with my typing.

 

[Denis]

can someone show me how to solve formulas with exponential like this one 100,000(1+x)^20=329,575

RULE: if x^p = y then x = y^(1/p)

100,000(1 + x)^20 = 329,575
(1 + x)^20 = 3.29575
1 + x = 3.29575^(1/20)
x = 3.29575^(1/20) - 1
x = .06144553139717... (ok, LookAgain?!)

 

[mmm4444bot]

got lazy with my typing

Go to the corner.