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Question 998682: solve for x
x= 6/8 (log(base6) sqrt(108) - 1 log(base6) (2/sqrt(3))+ 2 log (base6) 9 + log(base6) 64)
I am not sure how to find X, I would appreciate if someone can help me. I will provide my step.
x=6/8 (log(base6) sqrt(108) - 1 log(base6) (2/sqrt(3))+ 2 log (base6) 9 + log(base6) 64)
=6/8 (log(base6) sqrt(108)/ (2/sqrt3) + log(base 6) 9^2 + log(base6) 64)
=6/8 (log(base6) sqrt(108) / (2/sqrt3) + log(base 6) (81)(64)
=6/8 (log(base6) 9 + log(base6) 5184)

=6/8 (log(base6) (9)(5184)
=6/8 (log(base6) 46656)
= log(base6) 34992

Found 2 solutions by greenestamps, MathTherapy:
Answer by greenestamps(13367)   (Show Source):
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solve for x
x= 6/8 (log(base6) sqrt(108) - 1 log(base6) (2/sqrt(3))+ 2 log (base6) 9 + log(base6) 64)
I am not sure how to find X, I would appreciate if someone can help me. I will provide my step.
x=6/8 (log(base6) sqrt(108) - 1 log(base6) (2/sqrt(3))+ 2 log (base6) 9 + log(base6) 64)
=6/8 (log(base6) sqrt(108)/ (2/sqrt3) + log(base 6) 9^2 + log(base6) 64)
=6/8 (log(base6) sqrt(108) / (2/sqrt3) + log(base 6) (81)(64)
=6/8 (log(base6) 9 + log(base6) 5184)
=6/8 (log(base6) (9)(5184)
=6/8 (log(base6) 46656)
= log(base6) 34992

------------------------------------------------------------------

Your work is fine up to the last step.

It will be easier to find the answer if you leave the expression in factored form instead of multiplying all the numbers together.

The entire expression is of the form

(6/8) (... extended expression in a form equivalent to "log (base6) of A")

Let's drop all the "log (base 6)" notations and just look at the "A". Using basic log rules, A is as follows:

Performing basic arithmetic WITHOUT multiplying all the numbers together....

And now remembering that we are looking for the value of the log base 6 of this number....

Finally, then....

ANSWER: 9/2

Answer by MathTherapy(10858)   (Show Source):
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solve for x x= 6/8 (log(base6) sqrt(108) - 1 log(base6) (2/sqrt(3))+ 2 log (base6) 9 + log(base6) 64) I am not sure how to find X, I would appreciate if someone can help me. I will provide my step. x=6/8 (log(base6) sqrt(108) - 1 log(base6) (2/sqrt(3))+ 2 log (base6) 9 + log(base6) 64) =6/8 (log(base6) sqrt(108)/ (2/sqrt3) + log(base 6) 9^2 + log(base6) 64) =6/8 (log(base6) sqrt(108) / (2/sqrt3) + log(base 6) (81)(64) =6/8 (log(base6) 9 + log(base6) 5184) =6/8 (log(base6) (9)(5184) =6/8 (log(base6) 46656) = log(base6) 34992 ******************* ---- Applying

Question 719241: Log x - log 4= -2
Found 2 solutions by josgarithmetic, MathTherapy:
Answer by josgarithmetic(39838)   (Show Source):
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Log x - log 4= -2
----

assuming common logs

Answer by MathTherapy(10858)   (Show Source):
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Log x - log 4= -2 ***************** log (x) - log (4) = - 2 -------- Applying = --- Converting from LOGARITHMIC to EXPONENTIAL form --- Cross-multiplying

Question 1157957: The population of Plano, Texas, which follows the exponential growth model, increased from
222,030 in 2000 to 259,841 in 2010. (Source: U.S. Census Bureau, www.census.gov)
Work is required!
a. Find the exponential growth rate, k. Give the exact rate.

b. Write the exponential growth function that models the population after t years.
c. What is the projected population in 2020?
d. How long should it take the population to double?
Answer by KMST(5396)   (Show Source):
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Exponential functions are functions where the variable is in the exponent, such as .
Exponential functions of the form where and are constants
are used to model exponential growth and exponential decay.
Let's call our function for population.
Let be the number of years after 2000.
The function is .
For the year 2000, and , and
--> -->
For the year 2010, , , and
--> --> -->
That is the exact value.
or .

We could write the growth function as
, or
, or
, or ,
or maybe use the approximate value for k, and write it as

We can calculate the projected population in 2010 and 2020, using the equations found above.
For 2010, . Substituting that value into , we get
(rounded to whole number).
However using , we get
(rounded to whole number).
To match all 6 digits in the population for 2010, we need the exact value of , or at least a better approximation and we need to carry more digits through the calculations.
For 2020 , substituting it into we get
.
Rounding to whole numbers, we get that the projected population in 2020 is .
If we use the more accurate , we get
.
Rounding to whole numbers, we get that the projected population in 2020 is .

Using the equation with the exact value of , , we could set and solve for .
We get
-->--> (rounded to 3 decimal places).
Rounding to whole numbers, it takes years for the population to double at the rate observed between 2000 and 2010.

Question 47806: what is 3^500?
Found 2 solutions by greenestamps, ikleyn:
Answer by greenestamps(13367)   (Show Source):
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You can find exact values of large integers like this using the PARI online calculator. Go to

pari.math.u-bordeaux.fr

How to find the calculator when you get to that web site apparently depends on your browser.

ANSWER: 3^500 =

36360291795869936842385267079543319118023385026001623040346035832580600191583895484198508262979388783308179702534403855752855931517013066142992430916562025780021771247847643450125342836565813209972590371590152578728008385990139795377610001

Answer by ikleyn(53937)   (Show Source):
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.
.
what is 3^500?
~~~~~~~~~~~~~~~~~~~~~~

        In his post, @consc198 responds "no simplification!".
        But the problem does not ask to simplify. It asks to evaluate (or to estimate) the value.
        @consc198 is so ignorant of mathematics that does not understand the meaning of the request.

        Meanwhile, the solution is standard and straightforward.

Take decimal logarithm of . = 500*log(3) = 500*0.477121255 = 238.5606274. It tells us that is multiplied by = = , approximately. ANSWER. is approximately .

Solved.

This is so important skill that every educated person
should have it in his or her hands, as well as in his or her mind.

Question 473939: How do i solve this log5^7+1/2log5^4=log5^x
Answer by MathTherapy(10858)   (Show Source):
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How do i solve this log5^7+1/2log5^4=log5^x ******************************************* The other person got x = 9, but this author disagrees, as this most likely is: --- Applying = ---- Applying = 14 = x --- LOG bases are same, so LOG arguments are EQUIVALENT

Question 46884: Solve the equation 3log5 x - log5 4 = log5 16
Answer by MathTherapy(10858)   (Show Source):
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Solve the equation 3log5 x - log5 4 = log5 16 ********************************************* Everything that the other person, consc198(59), did, was WRONG. He/she contends that: "3log5 x - log5 x 4 = log5 x 16 2.096910013 x -0.698970004 x 4 = 0.698970004 x 16 2.096910013 x -2.795880016 = 11.18352006 -5.862708801 = 11.18352006 11.18352006 + 5.862708801 Answer = 17.04622886" ******* This is CERTAINLY NOT the way to solve a LOGARITHMIC equation. Furthermore, the answer is CERTAINLY NOT 17.04622886. Instead, it's done PRECISELY as follows: ---- Applying = ---- Applying ---- If , then c = d ------ Cross-multiplying --- Taking the CUBE ROOT of each side x = 4

Question 289033: Hello,
I am having trouble with this question - could you please provide me with the steps, and solution?
If the half-life of Carbon 14 is 5730 years, how muc will remain of an initial 3 gms after 1000 years?
Thanks!
Found 3 solutions by greenestamps, josgarithmetic, ikleyn:
Answer by greenestamps(13367)   (Show Source):
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Using the definition of half life, the amount remaining after half-lives is , where is the initial amount.

In this problem, the number of half-lives is .

With an initial amount of 3 grams, the amount remaining after that many half-lives is

= 2.658186689....

Remember that radioactive decay is a statistical process; that mathematical answer is only approximately correct

Answer by josgarithmetic(39838)   (Show Source):
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If model is of the form
then for the given half-life

To use the formula and the other given information

Answer by ikleyn(53937)   (Show Source):
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.

If to keep four decimals after the decimal point, then
the correct answer is 2.6582 grams instead of 2.6581 in the post by @mananth.

Question 416707: Please help me solve this equation! log2 x+ log2 (x+1)=1
Found 2 solutions by ikleyn, MathTherapy:
Answer by ikleyn(53937)   (Show Source):
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.
Please help me solve this equation! log2 x+ log2 (x+1)=1
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

        Below is my complete correct solution

log2(x) + log2(x+1) = 1

The domain for this equation is the set of all real positive numbers x > 0.

In this domain, the given equation is equivalent to this one

log2(x*(x+1)) = 1

Simplify and find x

x(x+1) = 2^1,
x^2+x-2 = 0
(x+2)(x-1) = 0
x = -2, 1

x = -2 is an extraneous solution out the domain, so we disregard x = -2
(since logarithm of the negative argument is not defined).

Thus, x = 1 is the unique real solution to the original equation.         <<<---===   ANSWER

Solved.

Answer by MathTherapy(10858)   (Show Source):
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Please help me solve this equation! log2 x+ log2 (x+1)=1 *********************************************************** , with x > 0 The other person's claim, that - 2 is a solution to this equation, is FALSE. 1 is, though, since it's > 0!!

Question 62416: A population is increasing at a rate of 0.6% each year. The population is currently 3000. Find a function p(t)that gives the size of the population t years from now.

Answer by ikleyn(53937)   (Show Source):
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.
A population is increasing at a rate of 0.6% each year. The population is currently 3000.
find the function p(t)that gives the size of the population t years from now.
~~~~~~~~~~~~~~~~~~~~~~~~~

        The solution in the post by @jai_kos is incorrect.

The function p(t) that gives the size of the population t years from now is p(t) = = . ANSWER

Solved correctly.

Question 62314: A population is increasing at a rate of 0.6% each year. The population is currently 3000. find the function p(t)that gives the size of the population t years from now.

Answer by ikleyn(53937)   (Show Source):
You can put this solution on YOUR website!
.
A population is increasing at a rate of 0.6% each year. The population is currently 3000.
find the function p(t)that gives the size of the population t years from now.
~~~~~~~~~~~~~~~~~~~~~~~~~

        The solution in the post by @jai_kos is incorrect.

The function p(t) that gives the size of the population t years from now is p(t) = = . ANSWER

Solved correctly.

Question 1078473:
Solve. Justify your work
The number above the square root of 8 is a one
Answer by MathTherapy(10858)   (Show Source):
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Solve. Justify your work The number above the square root of 8 is a one ********************************************** The other person's answer, is WRONG!! _________ ____________ The EXPONENTS are the same, so equating the BASES, x = 3

Question 1149061: Abby opened a retirement account with 9% APR and initial deposit of $8,000 compounded monthly.
(a) Find the exponential function that models the value of Abby's retirement account after t years.
(b) How long will it take to double the initial value? Estimate your answer to the nearest year.
(c) How long will it take to triple the initial value? Estimate your answer to the nearest year.
(d) How long will it take for the value of the account to reach 7 times the value of the account after 2 years?
Answer by MathTherapy(10858)   (Show Source):
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Abby opened a retirement account with 9% APR and initial deposit of $8,000 compounded monthly. (a) Find the exponential function that models the value of Abby's retirement account after t years. (b) How long will it take to double the initial value? Estimate your answer to the nearest year. (c) How long will it take to triple the initial value? Estimate your answer to the nearest year. (d) How long will it take for the value of the account to reach 7 times the value of the account after 2 years? **************************************** (a) Find the exponential function that models the value of Abby's retirement account after t years. Future-value-of-$1 formula: , with = Future Value (Unknown, in this case) = Principal/Initial Deposit ($8,000, in this case) = Interest rate, as a decimal (9%, or .09, in this case) = Number of ANNUAL compounding periods (monthly, or 12, in this case) = Time Principal/Initial Deposit has been invested, in YEARS (t, in this case) ----- Substituting $8,000 for P, .09 for i, and 12 for m <=== Exponential function that models Abby's account-value, after t years ========================================================= (b) How long will it take to double the initial value? Estimate your answer to the nearest year. ----- Substituting $16,000 for A, $8,000 for P, .09 for i, and 12 for m ----- Converting to LOGARITHMIC form Time it'll take to double the initial value, or = 7.730480505, or approximately 8 years. ======================================================== (c) How long will it take to triple the initial value? Estimate your answer to the nearest year. ----- Substituting $24,000 for A, $8,000 for P, .09 for i, and 12 for m ----- Converting to LOGARITHMIC form Time it'll take to triple the initial value, or = 12.25252171, or approximately 13 years. ** Notice that although 12.25252171 rounds off to 12 (nearest integer), the $8,000 investment, at the 12-year juncture, will be less than TRIPLE, or less than $24,000 (3 * $8,000). This is why it's necessary to ROUND UP to year 13, at which time, the $8,000 initial deposit will surpass the triple-value, $24,000. =============================================================== How long will it take for the value of the account to reach 7 times the value of the account after 2 years? <=== Exponential function that models Abby's account-value, after t years ---- Substituting 2 for t Value of account after 2 years, or A = 9571.308235, or approximately $9,572. Seven (7) times the 2-year account value = 7(9,572) = $67,004. Now, when will the initial deposit ($8,000) increase to $67,004? ----- Substituting $67,004 for A, $8,000 for P, .09 for i, and 12 for m ----- Converting to LOGARITHMIC form It'll take = 23.70300853, or approximately 24 years for the $8,000 initial-deposit to increase to $67,004 (7 times its 2-year value, or 7 times appr. $9,572).

Question 1095894: Solve for A if f(x)=-Ax^2+Bx-8, f(-5)=-123 and g(x)=4x+B, g(2)=11
Answer by MathTherapy(10858)   (Show Source):
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Solve for A if f(x)=-Ax^2+Bx-8, f(-5)=-123 and g(x)=4x+B, g(2)=11 ***************************************************************** Solve for A if and f(- 5) = - 25A - 5B - 8 - 123 = - 25A - 5B - 8 25A + 5B = 115 ------ eq (i) g(x) = 4x + B g(2) = 4(2) + B 11 = 8 + B 3 = B 25A + 5B = 115 ---- eq (i) 25A + 5(3) = 115 ---- Substituting 3 for B in eq (i) 25A + 15 = 115 25A = 100 (A, B) = (4, 3)

Question 1113133: Show that log base a (a^2 - x^2) = 2 + log base a (1 - x^2/a^2)
Answer by MathTherapy(10858)   (Show Source):
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Show that log base a (a^2 - x^2) = 2 + log base a (1 - x^2/a^2) *************************************************************** Let's focus on the R.H.S ---- Converting 2 to --- Applying = ---- Distributing on R.H.S. L.H.S. = R.H.S.

Question 1104549: Please help me solve this word problem: in 1997, there were a total of 1,316,999 bankruptcies filed under the Bankruptcy Reform Act. The model for the number of bankruptcies filed is B(t)=0.798*1.164^t, where t is the number of years since 1994 and B is the number of bankruptcies filed in terms of millions (administrative office in US courts, statistical tables for the federal judiciary). In what year is it predicted that 12 million bankruptcies will be filed?
Found 3 solutions by ikleyn, josgarithmetic, MathTherapy:
Answer by ikleyn(53937)   (Show Source):
You can put this solution on YOUR website!
.
Please help me solve this word problem:
in 1997, there were a total of 1,316,999 bankruptcies filed under the Bankruptcy Reform Act.
The model for the number of bankruptcies filed is B(t)=0.798*1.164^t, where t is the number of years since 1994
and B is the number of bankruptcies filed in terms of millions (administrative office in US courts,
statistical tables for the federal judiciary). In what year is it predicted that 12 million bankruptcies will be filed?
~~~~~~~~~~~~~~~~~~~~~~~~~

Been a curious person, I asked Google: "What is the maximum number of bankrupts per year in the US history ?"

Below is the Google answer

The maximum number of bankruptcy filings in US history occurred in 2005, with over two million cases filed.
This record spike was driven by a rush to file before the implementation of the Bankruptcy Abuse Prevention
and Consumer Protection Act (BAPCPA) in October of that year.
Key Data Points:
Historical Peak: 2005 saw > 2 million total (individual and business) cases.
Pre-2005 Surge: 1 in 55 households filed for bankruptcy in 2005.
Recent Corporate Highs: In terms of large corporate bankruptcies, 2024 saw a 14-year high with 694 companies filing, according to data from LPL Financial.
Record Failure: The largest bankruptcy by assets in US history was Lehman Brothers in 2008 ($691 billion), notes
Statista.

///////////////////////////

So, based on this data, I think that the proposed model in the post
is NOT adequate, has NO any relation to reality, is pure imagination and is fundamentally wrong.

Answer by josgarithmetic(39838)   (Show Source):
You can put this solution on YOUR website!
B is in the millions. Just plugging into the formula,

. Compute this how you want.

t looks like about 17.8 years.

Answer by MathTherapy(10858)   (Show Source):
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Please help me solve this word problem: in 1997, there were a total of 1,316,999 bankruptcies filed under the Bankruptcy Reform Act. The model for the number of bankruptcies filed is B(t)=0.798*1.164^t, where t is the number of years since 1994 and B is the number of bankruptcies filed in terms of millions (administrative office in US courts, statistical tables for the federal judiciary). In what year is it predicted that 12 million bankruptcies will be filed?
*******************************************************************************

---- Substituting 12 for B(t), the number of bankruptcies, in MILLIONS

-- Converting to LOGARITHMIC form
t = 17.8488, approximately
With t representing TIME, in years, since 1994, we get the year that 12M bankruptcies
will be predicted to be filed as: 1994 + 17.8488, or 2011.8488, which is in the year 2011.

Question 822027: log3 15 − log3 40 + log3 648
Answer by MathTherapy(10858)   (Show Source):
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log3 15 − log3 40 + log3 648 ********************************** I take it you need this SIMPLIFIED!! ----- Applying = = = = = = = 5

Question 734837: solve log3(x-3) + log3(x+2) = log3 6
Answer by MathTherapy(10858)   (Show Source):
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solve log3(x-3) + log3(x+2) = log3 6 ************************************ The other person's statement that if "all log bases are the same (namely 3)", then: x-3 + x+2 = 6 2x - 1 = 6 and 2x = 7 and x = 3.5, is a FALLACY! He couldn't be more WRONG!! If he'd checked, he would've seen that 3.5 doesn't: make sense make the equation TRUE! This log equation starts with 2 log arguments: a1 (x - 3), and a2 (x + 2), the smaller being log argument a1, x - 3. Argument "a1" "tells" one that x - 3 MUST be > 0. So, x - 3 > 0_____x > 3 We then get: , with x > 3 ----- Applying = (x - 3)(x + 2) = 6 ------ Applying c = d, if (x - 4)(x + 3) = 0 x - 4 = 0 OR x + 3 = 0 x = 4 OR x = - 3 As stated above, x MUST be > 3, so ONLY x = 4 is ACCEPTABLE. This makes x = - 3, an EXTRANEOUS solution to this equation, and is therefore IGNORED/REJECTED.

Question 977752: If log{{x+y}/3}=1/2{logx+log Y}, then find the value of x/y+y/x?
Answer by MathTherapy(10858)   (Show Source):
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If log{{x+y}/3}=1/2{logx+log Y}, then find the value of x/y+y/x? ================================================================ What the other person who responded did and said, doesn't make any sense, at all, to this author. Plus, nowhere in his response, is there an answer to the question: What's the value of ? ---- Multiplying by 2 ---- Applying c = d, if ---- FOILing ---- Dividing each expression on left-side by xy

Question 974318: If then = ?
Answer by MathTherapy(10858)   (Show Source):
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If then = ? ==================================================================== I don't know what convoluted stuff that other person who responded wrote, but all that MUMBO JUMBO seems, in my opinion, to have absolutley nothing to do with the expression to be evaluated! If then = ? ---- Substituting 7xy for ----- Taking the square root of each side = ? --- Substituting ONLY the POSITIVE value (), for x + y ----- Applying = = , or

Question 280462: This one looked easy but it's got me confused please help and thank you.
Use the laws of logarithms to express log base2 6 - log base 2 3 + 2 log base 2 sqrt8 as a single logarithm then evaluate. I got this far log base 2^6 - log base2^3 +2 log base 2 sqrt 8 =log base 2 6/3 sqrt8^2??? help
Found 2 solutions by greenestamps, MathTherapy:
Answer by greenestamps(13367)   (Show Source):
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You have done the harder part of the problem, using the laws of logarithms to write the given expression as a single logarithm. The rest of the problem is simple arithmetic, plus an application of the definition of logarithms:

So you have

which is 4, because

ANSWER: 4

Answer by MathTherapy(10858)   (Show Source):
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This one looked easy but it's got me confused please help and thank you. Use the laws of logarithms to express log base2 6 - log base 2 3 + 2 log base 2 sqrt8 as a single logarithm then evaluate. I got this far log base 2^6 - log base2^3 +2 log base 2 sqrt 8 =log base 2 6/3 sqrt8^2??? help ============================================================

Question 718152: Please help me solve for A and B in this exponential decay model y= ae^-bx. I have the following points: when x=0.6, y=1000 and when x=1.8, y=1.
I have done the following but I am not convienced it is the right path...
y=ae-bx therefor e-bx=y/a ... take log e of both sides ... loge e-bx = loge y/a
but logee=1 therefor -bx=loge y/a ... move the x over to get -b=x loge y/a
then enter values for x and y from example 1... -b= (0.6) loge(1000/a)
Is this the correct path to take for solving this?? Any advice would be greafull appreciated. Thank you.
Found 2 solutions by greenestamps, MathTherapy:
Answer by greenestamps(13367)   (Show Source):
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The function has the form

The two given data points are (0.6,1000) and (1.8,1), giving us these two equations:

Dividing the first equation by the second will eliminate a, giving us an equation we can solve to find b.

A calculator then gives the value of b to several decimal places as 5.75646...

Use that value of b in the second equation to get an equation we can solve to find the value of a.

Using a calculator then gives the value of a to several decimal places as 31622.7766....

ANSWERS (to several decimal places):
a = 31622.7766
b = 5.75646

Graphing this function on a graphing calculator confirms the two given data points (0.6,1000) and (1.8,1).

Found using a different path than that used by the other tutor, this matches his answers -- except that currently he has a typo in the leading digit of the value for a.

Answer by MathTherapy(10858)   (Show Source):
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Please help me solve for A and B in this exponential decay model y= ae^-bx. I have the following points: when x=0.6, y=1000 and when x=1.8, y=1. I have done the following but I am not convienced it is the right path... y=ae-bx therefor e-bx=y/a ... take log e of both sides ... loge e-bx = loge y/a but logee=1 therefor -bx=loge y/a ... move the x over to get -b=x loge y/a then enter values for x and y from example 1... -b= (0.6) loge(1000/a) Is this the correct path to take for solving this?? Any advice would be greafull appreciated. Thank you. ======================================================================================================== I've chosen to show you the CORRECT path, instead of looking through your work to identify your error(s), if any! , if: , if: Equating the 2 "a" values, we get: ---- Cross-multiplying --- Dividing both sides by 1.2b = ln (1,000) --- Converting to LOGARITHMIC (natural) form = 5.76, approximately. You can now substitute this value for b in either , or to determine the value of "a." If/When you do, you should get the value of a as approximately 31,622.8, to the nearest tenth. @GREENESTAMPS is CORRECT! An INCORRECT value of 51,622.8 for "a" was entered previously, although the correct value, 31,622.8 was calculated. This has now been corrected. Thx for pointing out the error.

Question 942780: Solve for x.
a.) log sub2(log sub4{log sub3[log sub2 x]}) = -1
b.) log sub2(x+4) + log sub2(x-2) - log sub2(x-6) = 0
Answer by MathTherapy(10858)   (Show Source):
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a.) log sub2(log sub4{log sub3[log sub2 x]}) = -1 b.) log sub2(x+4) + log sub2(x-2) - log sub2(x-6) = 0 a) PAINSTAKINGLY LONG solution by the other person. b) Other person's solutions are EXTRANEOUS, and therefore, UNACCEPTABLE!!! b.) log sub2(x+4) + log sub2(x-2) - log sub2(x-6) = 0 Smallest log: (x - 6), so x - 6 > 0. Therefore, x > 6. We then have: , with x > 6. The 2 solutions the other person got, x = - 2, and x = 1 are < 6, NOT > 6, obviously. This makes them EXTRANEOUS, thereby leading to this equation having NO SOLUTIONS!

Question 959061: how do you solve log base 4 (log base 3 (log base 2 of X)) = 0??

Answer by MathTherapy(10858)   (Show Source):
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how do you solve log base 4 (log base 3 (log base 2 of X)) = 0?? DON'T know what the other person who responded did, but what was seems quite CONFUSING, and lacked a CONCISE answer.

Question 31666: Oh this logarithm problems are killing me :)
Can you help! Write hte logarithm of a single expression:
log[b]6x+4(log[b]x-log[b]y)
(A) log[b] 6x^5/y^4
(B) log [b] 24x^2/y
(C) log [b] 10x/4y
(D) none of these.
Do we even consider the "B" in this expression?
Answer by MathTherapy(10858)   (Show Source):
You can put this solution on YOUR website!

Oh this logarithm problems are killing me :) Can you help! Write hte logarithm of a single expression: log[b]6x+4(log[b]x-log[b]y) (A) log[b] 6x^5/y^4 (B) log [b] 24x^2/y (C) log [b] 10x/4y (D) none of these. Do we even consider the "B" in this expression? log[b]6x+4(log[b]x-log[b]y) --- Distributing ---- Applying = --- Applying and Do we even consider the "B" in this expression? Don't even consider B or C, as they obviously DO NOT MAKE SENSE. CHOICE D, obviously, is NOT correct either.

Question 738696: Log base 9 of 3 to the 3x-4=x. How do I solve for x
Answer by MathTherapy(10858)   (Show Source):
You can put this solution on YOUR website!

Log base 9 of 3 to the 3x-4=x. How do I solve for x This is easier than it "looks"!! ----- Converting to EXPONENTIAL form 3x - 4 = 2x ----- Exponents are equated, since bases are equal - 4 = 2x - 3x - 4 = - x

Question 452859: what does x equal?
log 10,000=x
Answer by MathTherapy(10858)   (Show Source):
You can put this solution on YOUR website!

what does x equal? log 10,000=x log 10,000 = x ---- Converting to EXPONENTIAL form ---- Converting 10,000 to x = 4 ---- Equating EXPONENTS, since BASES are equal

Question 47675: Logarithms:
a) Using a calculator, find log 1000 where log means log to the base of 10.
Answer:

b) Most calculators have 2 different logs on them: log, which is base 10, and ln, which is base e. In computer science, digital computers are based on the binary numbering system which means that there are only 2 numbers available to the computer, 0 and 1. When a computer scientist needs a logarithm, he needs a log to base 2 which is not on any calculator. To find the log of a number to any base, we can use a conversion formula as shown here:

Using this formula,a=log a/log b, find log2 1000 . Round your answer to the hundredth's place.
Answer:
Show work in this space.

Answer by MathTherapy(10858)   (Show Source):
You can put this solution on YOUR website!

Logarithms: a) Using a calculator, find log 1000 where log means log to the base of 10. Answer: b) Most calculators have 2 different logs on them: log, which is base 10, and ln, which is base e. In computer science, digital computers are based on the binary numbering system which means that there are only 2 numbers available to the computer, 0 and 1. When a computer scientist needs a logarithm, he needs a log to base 2 which is not on any calculator. To find the log of a number to any base, we can use a conversion formula as shown here: Using this formula,a=log a/log b, find log2 1000 . Round your answer to the hundredth's place. Answer: Show work in this space. b) = = = = = = 9.97, approximately

Question 454898: Can you show me step by step how to solve this equation?
You deposit $1000 into a savings account in which the intrest is compounded anuualy at a rate of 5%.

What is the pricipal after 5 years?
Answer by ikleyn(53937)   (Show Source):
You can put this solution on YOUR website!
.
Can you show me step by step how to solve this equation?
You deposit $1000 into a savings account in which the intrest is compounded anuualy at a rate of 5%.

What is the pricipal after 5 years?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

The question in the post is posed incorrectly and shows
that a person who created/composed the problem did not know the terminology of the subject.

A correct question should ask about the Future value of the account in 5 years.

The principal is and remains unchangeable.

Question 253889: beryllium-11 decomposes into boron-11 with a half life of 13.8 seconds. How long will it take 240 g of beryllium-11 to decompose into 7.5g of beryllium-11?
based on the forumula provided in the text book C(t)=C(0)2e^-t/h
where C(0) = 240, C(t)=7.5 ,h=13.8
7.5/240=2e^-t/13.8
this was a s far as i got
Answer by ikleyn(53937)   (Show Source):
You can put this solution on YOUR website!
.
Beryllium-11 decomposes into boron-11 with a half life of 13.8 seconds.
How long will it take 240 g of beryllium-11 to decompose into 7.5 g of berlyium-11?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

This problem is very special and very specific.

For the complete solution, there is NO NEED to write many-store complicated exponential functions.

There is a special method and a special approach/reasoning, and this problem
is SPECIALLY CREATED for you to learn this special solution method from me.

Notice that the ratio      is   32.

        It means that 5 (five) half-lives happened in this decay process
        from 240 gram of Beryllium-11 to 7.5 grams of Beryllium-11, because    = 32.

5 half-lives is 5 times 13.8 seconds, or 5 * 13.8 = 69 seconds.

ANSWER.   It will take 69 seconds for 240 grams of   Beryllium-11 to decompose into 7.5 grams of Beryllium-11.

*************************************************************

        Solved and explained in a way how it SHOULD be done
                    and how it is EXPECTED to be done.

*************************************************************

Question 253874: beryllium-11 decomposes into boron-11 with a half life of 13.8 seconds. How long will it take 240 g of beryllium-11 to decompose into 7.5g of beryllium-11?
based on the forumula provided in the text book C(t)=C(0)2e^-t/h
where C(0) = 240, C(t)=7.5 ,h=13.8
7.5/240=2e^-t/13.8
this was a s far as i got
Answer by ikleyn(53937)   (Show Source):
You can put this solution on YOUR website!
.
Beryllium-11 decomposes into boron-11 with a half life of 13.8 seconds.
How long will it take 240 g of beryllium-11 to decompose into 7.5 g of berlyium-11?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

This problem is very special and very specific.

For the complete solution, there is NO NEED to write many-store complicated exponential functions.

There is a special method and a special approach/reasoning, and this problem
is SPECIALLY CREATED for you to learn this special solution method from me.

Notice that the ratio      is   32.

        It means that 5 (five) half-lives happened in this decay process
        from 240 gram of Beryllium-11 to 7.5 grams of Beryllium-11, because    = 32.

5 half-lives is 5 times 13.8 seconds, or 5 * 13.8 = 69 seconds.

ANSWER.   It will take 69 seconds for 240 grams of   Beryllium-11 to decompose into 7.5 grams of Beryllium-11.

*************************************************************

        Solved and explained in a way how it SHOULD be done
                    and how it is EXPECTED to be done.

*************************************************************

Question 730115: Berlyium-11 decomposes into boron-11 with a half life of 13.8 seconds. How long will it take 2400g of berlyium-11 to decompose into 75g of berlyium-11?

Thats the question, and I tried solving it but im stuck. This is what I have so far:
75=2400 x 2^-t/13.8
0.01325=27.6 (-t)
Answer by ikleyn(53937)   (Show Source):
You can put this solution on YOUR website!
.
Beryllium-11 decomposes into boron-11 with a half life of 13.8 seconds.
How long will it take 2400g of beryllium-11 to decompose into 75g of berlyium-11?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

First, from chemical and physical points of views, your post was written terrifically.
I mean the name of Beryllium. So I re-wrote it in correct form.

Next, this problem is very special and very specific.

For the complete solution, there is NO NEED to write many-store complicated exponential functions.

There is a special method and a special approach/reasoning, and this problem
is SPECIALLY CREATED for you to learn this special solution method from me.

Notice that the ratio      is   32.

        It means that 5 (five) half-lives happened in this decay process
        from 2400 gram of Beryllium-11 to 75 grams of Beryllium-11, because    = 32.

5 half-lives is 5 times 13.8 seconds, or 5 * 13.8 = 69 seconds.

ANSWER.   It will take 69 seconds for 2400 grams of   Beryllium-11 to decompose to 75 grams of Beryllium-11.

*************************************************************

        Solved and explained in a way how it SHOULD be done
                    and how it is EXPECTED to be done.

*************************************************************

Question 742195: One pipe can fill a tank in 20 minutes, while
another takes 30 minutes to fill the same tank. How long would it take the two pipes together to fill the tank?

Found 4 solutions by MathTherapy, josgarithmetic, greenestamps, ikleyn:
Answer by MathTherapy(10858)   (Show Source):
You can put this solution on YOUR website!

One pipe can fill a tank in 20 minutes, while another takes 30 minutes to fill the same tank. How long would it take the two pipes together to fill the tank? Answer is RIGHT HERE, below.....Question 45662: Hi again, I need some assistance with this mixture problem: It takes a pipe 10 minutes less than another one to fill a tank of water. Both pipes together can fill the tank in 12 minutes. How long will it take each one to fill the tank separately? I am having trouble on how to start setting the equation up. Thanks again, Lou

Answer by josgarithmetic(39838)   (Show Source):
You can put this solution on YOUR website!

RATE TIME TANKS ONE PIPE 1/20 20 1 ANOTHER PIPE 1/30 30 1

If the two pipes are used together for filling the one tank, then

---------------twelve minutes

Answer by greenestamps(13367)   (Show Source):
You can put this solution on YOUR website!

Here is a different way of solving a problem like this.

Consider the least common multiple of the two given times, which is 60 minutes.

In 60 minutes, the first pipe could fill the tank 60/20 = 3 times.
In 60 minutes, the second pipe could fill the tank 60/30 = 2 times.
So in 60 minutes together the two pipes could fill the tank 3+2 = 5 times.
So the time it takes the two pipes to fill the tank one time is 60/5 = 12 minutes.

ANSWER: 12 minutes

NOTE: Both this method and the method shown by the other tutor can be used to solve similar problems in which one pipe is trying to fill the tank while another pipe is trying to drain the tank.

Answer by ikleyn(53937)   (Show Source):
You can put this solution on YOUR website!
.
One pipe can fill a tank in 20 minutes, while another takes 30 minutes to fill the same tank.
How long would it take the two pipes together to fill the tank?
~~~~~~~~~~~~~~~~~~~~~~~~~

First pipe fills 1/20 of the tank volume per minute when works alone. Second pipe fills 1/30 of the tank volume per minute when works alone. Working together, both pipes fill + = + = = of the tank volume per minute. Hence, it will take 12 minutes for both pipes to fill the tank. ANSWER

Solved.

The answer in the post by @lynnlo is incorrect.