simplify log5[(4x+1)^3 ?(4x+y^2)/(5x+1)^10], solve (e^5x)+(

[SMS]

the first question is
Simplify, log5[(4x+1)^3 ?(4x + y^2) / (5x + 1)^10]

and

Solve, (e^5x) + (e^-5x) / 2 = 3

thanks for any help in advance

 

[Deleted member 4993]

Re: need help with some log questions

the first question is
Simplify, log5[(4x+1)^3 ?(4x + y^2) / (5x + 1)^10]

does the problem above have log(base 5)?

in any case, use the properties of logs - that says

\(\displaystyle log_m(a\cdot b) \, = \, log_m(a) \, + \, log_m(b)\)

and

\(\displaystyle log_m(a^b) \, = \, b\cdot log_m(a)\)

and

Solve, (e^5x) + (e^-5x) / 2 = 3

Hint: substitute

\(\displaystyle u \, = \, e^{5x}\)

this will give you a quadratic equation - and solve that. Then substitute back to solve for 'x'
thanks for any help in advance

Please show us your work, indicating exactly where you are stuck - so that we know where to begin to help you.

 

[SMS]

Re: need help with some log questions

I dont understand the whole problem, if by any chance you can do each step by step and show the result that would help a lot, and its log base 5, sorry for the confusion

 

[Deleted member 4993]

Re: need help with some log questions

For some worked out examples - go to:

http://www.purplemath.com/modules/logrules2.htm

then come back and show us your work for further help.

 

[stapel]

Simplify, log5[(4x+1)^3 ?(4x + y^2) / (5x + 1)^10]

As suggested earlier, please first learn how to manipulate logs expressions such as the above.

Solve, (e^5x) + (e^-5x) / 2 = 3

This is, in essence, a quadratic. The "trick" for solving this sort of exponential equation is to multiply through by something to get rid of the negative power. In this case, multiply through by e[sup:3efqqbpn]5x[/sup:3efqqbpn]. This gives you:

. . . . .\(\displaystyle \left(e^{5x}\right)^2\, +\, \frac{1}{2}\, =\, 3e^{5x}\)

(Note: In the above, I assumed that you meant the "2" to be under only the second term on the left-hand side, as you'd posted.)