Question 1

In number 5, where did the 1/2 come from. Why... Actually, it would be great if the whole problem is explained to me, but differently. It doesn't make sense to me.

Accepted Answer

I didn't use "1/2" and I don't think it's really necessary:

1) 4*(5^(2*x)) = 300

2) 5^(2*x) = 75  (At this point you can just take the log of both sides (see below) and that's where the 1/2 comes from), or:

3) (5^2)^x = 75 (laws of exponents)

4) 25^x = 75

5) x = log_25(75) (log form)

6) x = log75/log25 = 1.341 (change of base rule)

Or for 3) you could do:

3) log_5(5^(2*x)) = log_5(75)  (log form)

4) 2*x = log_5(75) (Here's where 1/2 comes in)

5) x = (1/2)log_5(75)

6) x = (1/2)(log75/log5) = 1.341

Question 2

while practicing a question came to my mind. It maybe stupid,but I am going to ask it anyway. we have a subtraction property(log a-log b=log a/b) and change of base property(log_a b=log b/log a). I can understand when it goes in the said direction.but when it is applied in reverse how do i differentiate eg. log a/log b how so i determine which property to apply.

Accepted Answer

Your question was a long time ago, but here is an answer for anyone that may be wondering.
The subtraction property says that log(a)-log(b) is equal to log(a/b).
The change of base property says that log_a(b) is equal to (log_x(b))/(log_x(a)).
So, in the subtraction property the division is within the log, while for the change of base property we are really dividing the answers from the two logs.
So, if we are given log(a/b) to expand, we can use the subtraction property; if we are to condense (log(b))/(log(a)), then we should use the change of base property. 
I had a little difficulty understanding how to tell the difference myself at first. Hopefully this explanation was written clearly enough.

Question 3

Can anybody help:
log_5 y+log_y 5= 6  ?

Accepted Answer

Really nice question! You need to know the properties of logarithms in order to solve this problem. 
It would be too tedious to type out the answer for you, so instead I linked it in this graph step by step.
https://www.desmos.com/calculator/9teeitvqjp

The answer is y = x=5^(3-2√2)

Hope this helps.

Question 4

Can someone walk me through the last challenge question? I see solving for each zero as they suggested, but am curious if one can expand it out as a quadratic to solve. When I try to do this, I keep coming out w incorrect solutions.

CHALLENGE QUESTION: Which of the following are solutions to (2^x-3)(2^x-4)=0

My approach is giving me {WHEN SETTING 2^x =y}:
y^2-7y+12=0

which then gives me a quadratic formula of;
7+or- sqrt(49-4(12)) all over 2

which results in values of 4 and 3.

I assume I am making an error, perhaps in setting the y or somewhere else regarding my treatment in expanding the equation, but I am not finding it; any help is appreciated.

Accepted Answer

In this setting we are solving for x, and how nice of them, to already give it to us factored out. When this quadratics is factored out this way, it means we are 1 step away from finding x itself. Both set of equations equal to 0 since we separated the into two individual parts.

2^x - 3 = 0
2^x = 3
x = log base 2 parentheses 3 is one of the answers.

`2^x - 4 = 0
2^x = 4
2 * 2 = 4 
x = 2 as the other answer.`

This is how it is sorted out. If you still have questions, please comment :)

Question 5

What exactly is going on when you have to change the base rule? I don't really follow that part of these equations

Accepted Answer

Did you watch the last section of videos "The Change of Base Formula for Logarithms"?  That really helped me, since I'd never heard of changing bases before.  Here's a link to the first video:
https://www.khanacademy.org/math/algebra2/exponential-and-logarithmic-functions/change-of-base-formula-for-logarithms/v/change-of-base-formula

Question 6

Hi I was wondering if you could help me with this question?
"Find the value of x in terms of a, where 3 log(base a)x=3+log(base a) 8"
Thanks

Accepted Answer

I believe that the short answer is: x = 2a.

Below is an explanation of how I got to that answer:

Start with: 3log_a (x) = 3 + log_a (8)
Multiply both sides by 1/3: log_a (x) = 1 + (1/3)log_a (8)
After using the power rule we get: log_a (x) = 1 + log_a (8^(1/3))
which simplifies to: log_a (x) = 1 + log_a (2)

Since 1 =  log_a (a), we can rewrite the equation as: 
log_a (x) = log_a (a) + log_a (2)
Next, using the Product Rule we get: log_a (x) = log_a (2a)

Finally, after canceling the log from both sides we get: x = 2a.

I hope that this is helpful.

Question 7

What if our equation has variables within the exponent. My equation looks like this, (1/9)^(2x+1) (27^x-2)=1

Accepted Answer

You need to get both sides to have a common base.
1)  Multiply both sides by 9^(2x+1).  This changes your equation into: 27^(x-2)=9^(2x+1)
2)  Factor 27 and 9.  27=3^3 and 9 = 3^2.  So, 3 can be used as a common base.  Replace 27 and 9 with their factors:
(3^3)^(x-2) = (3^2)^(2x+1)
3)  Multiply the exponents: 3^(3x-6) = 3^(4x+2)

Now, for these 2 sides to be equal, the exponents must be equal.  So, you can solve: 3x-6=4x+3 to find x.

Hope this helps.

Algebra 2

Course: Algebra 2 > Unit 8

Solving exponential equations using logarithms

Solving exponential equations of the form $a \cdot b^{x} = d$ ‍

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Solving exponential equations of the form $a \cdot b^{c x} = d$ ‍

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Algebra 2

Course: Algebra 2 > Unit 8

Solving exponential equations of the form a⋅bx=d‍

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Solving exponential equations of the form a⋅bcx=d‍

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Solving exponential equations of the form $a \cdot b^{x} = d$ ‍

Solving exponential equations of the form $a \cdot b^{c x} = d$ ‍