Solving Exponential & Logarithmic Functions

[tristatefabricatorsinc]

Hey,

Solve all for x

A) 100(1.02)^(x/4) = 200

B) log(x+25) = 1 + log(2x-7)

C) logx + log(x-21) = 2

I am very confused on how to solve these three problems... here are my thoughts on them below, I really need some assistance...

FOR A)
I do not know where to end or begin

FOR B)
When i tried to do this I subtracted log(2x-7) from the right side and brought it to the left to get.. log(x+25) - log(2x-7) = 1 and I tried to use the quotient rule

x+25
------- = 1
2x - 7

I am unsure if this is correct this far, but when I solve it, I do not get 5 for the answer which is what my book has. I am confused

FOR C)
I tried
log[(x)(x-21)] = 2
to get...
x^2 - 21x = 2

My book says the answer is 25, I am not sure if I have to do something with the 2 on the right or just subtract it from the right to get...

x^2 - 21x - 2 = 0

This is obviosuly not factorable, so Iwhen I used the quadratic form. it definitely did not come up with 25

Any help is greatly appreciated!

Thanks

 

[pka]

Assuming that log is the base 10:

Hey,
FOR B)
When i tried to do this I subtracted log(2x-7) from the right side and brought it to the left to get.. log(x+25) - log(2x-7) = 1 and I tried to use the quotient rule

x+25
------- = 10, this should be 10!
2x - 7

I am unsure if this is correct this far, but when I solve it, I do not get 5 for the answer which is what my book has. I am confused

FOR C)
I tried
log[(x)(x-21)] = 2
to get...
x^2 - 21x = 100, this should be 100!

My book says the answer is 25,
Thanks

 

[Denis]

A) 100(1.02)^(x/4) = 200

1.02^(x/4) = 200 / 100 = 2 : enough to get you going?