How Do You Simplify Logarithmic and Exponential Equations?

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The discussion focuses on simplifying logarithmic and exponential equations, with participants solving specific problems. The first problem involves simplifying ln(5th sqrt of y^2), leading to the correct answer being option A, 2/5 ln y. The second problem, log(base)5(x - 1) = -1, is solved correctly with the answer being C, 1.2. Another participant calculates the future value of $6000 at a 3.25% interest rate compounded quarterly for 5 years, arriving at the correct answer of D, 7054.05. The conversation highlights the importance of understanding logarithmic properties and compounding interest calculations.
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Here's the question:

Simplify the following as much as possible:

ln 5th Sqrt of y^2

a> 2/5 ln y

b> -3 + ln y

c> 3 + ln y

d> 5/2 ln y

A,B,C, or D, which is it?
 
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You can write the 5th sqrt of y^2 as y^(2/5) ... knowing that, it should be easy to get the right answer
 
I'm wanting to say "A", correct?
 
Yeap!

...
 
Thanks for the input, now for another, I'm pretty sure of my answer I just want to be sure. Here it is:

Solve the following equation

log(base)5(x - 1) = -1

a> 6

b> 0.2

c> 1.2

d> 0.8

My initial calculations (if they could be called that) came up with C, is that it?
 
Here's another while I'm at it:

Suppose you have $6000, how much does a return of 3.25% compounded quarterly for 5 years yield?

a> 6195

b> 7040.47

c> 6975

d> 7054.06

I'm wanting to say "B", but am unsure, any help?
 
***bump***
 
01 said:
Thanks for the input, now for another, I'm pretty sure of my answer I just want to be sure. Here it is:
Solve the following equation
log(base)5(x - 1) = -1
a> 6

b> 0.2

c> 1.2

d> 0.8

My initial calculations (if they could be called that) came up with C, is that it?

Well,it's right:
5^{\log_{5}(x-1)}=5^{-1}\Rightarrow x-1=\frac{1}{5}\Rightarrow x=\frac{6}{5}=1.2

Here's another while I'm at it:

Suppose you have $6000, how much does a return of 3.25% compounded quarterly for 5 years yield?

a> 6195

b> 7040.47

c> 6975

d> 7054.06

I'm wanting to say "B", but am unsure, any help?

So the interest is 3.25% per quarter of the year (trimester)??It means 13% per year.That means 780$ per year,which is 3900$ per 5 yrs.

If the interest is 3.25% per year,then it's 4 times less:975$ per 5 years,bringing it to a total of 6975$.Answer c).


In this case,what the hell means "compounded quarterly"??Capitalization once every three months?In this case,the rate per trimester is 3,25%/4=0.8125%.For the first trimester it would be 48.75$.Then for the second semester it would be:6048,75$*0.8125%=49.14609375$.So the total for 6 months would be:6097,89609375$.And for the third semester it would be 6097,89609375$*0.8125%=49.5454057617...$(the calcuator gave me only 10 decimals instead of 14).And the procedure would go on.
I don't know what the final result will be,but definitely it is weird.

Daniel.
 
Suppose you have $6000, how much does a return of 3.25% compounded quarterly for 5 years yield?

In this case,the rate per trimester is 3,25%/4=0.8125%

This is how do it. Develope a simple equation.
First Quarter: R1 = (6000 + \frac {0.8125}{100} 6000) = 6000(1+ \frac{0.8125}{100})


For the second Quarter: R2 = (R1 + \frac {0.8125}{100} R1) = R1(1+ \frac{0.8125}{100}) = 6000(1+ \frac{0.8125}{100})^2

There are 20 quarters in 5 years.
So the return after 5 years R20, is given by

R20 = 6000(1+ \frac{0.8125}{100})^{20}


This gives $ 7054.05 (d)


Regards,

Gamma.
 
  • #10
01 said:
Thanks for the input, now for another, I'm pretty sure of my answer I just want to be sure. Here it is:

Solve the following equation

log(base)5(x - 1) = -1

a> 6

b> 0.2

c> 1.2

d> 0.8

My initial calculations (if they could be called that) came up with C, is that it?

Another approach to this problem is to use the formula for changing the base, which is:

Log(base x)N=(Log(base 10)N)/(Log(base 10)x).

This can be very useful in any future quizzes or exams that you may have. This is also applicable for Ln (which is Log(base e)), e=2.7182...

Applying this formula to your original problem, the equation would look like this:

Log(base 5)(x-1)=-1
(Log(base 10)(x-1))/(log(base 10)5)=-1

Note: Log(base 10)=Log,
futher simplifying, we get

Log(x-1)=-1*Log5

Using properies of logarithms, we get Log(x-1)=Log(5^-1),
take the anti-logarithm of both sides we get, x-1=1/5

therefore answer is: x=1+1/5 or 1.2 :wink:

PS: Can someone help me with my thread (Supplementary angles for Spherical triangles)??
 
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