# Logarithmic problems.

• Apr 3rd 2007, 01:46 PM
Mulya66
Logarithmic problems.
Actual title of this thread is Logarithmic problems, sorry.

Anyway. I don't know how to use subscript, but any help/explanations would be grand. (note that in the first problem, there is nothing that is in subscript.)

"Evaluate the problem: log4 + 2log5"

"Solve the equation: log5x-3log52=log53"

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Also, "Solve the equation: logx32= 5"
But as for this one, I tried and got as far as this:
x^5 = 32
• Apr 3rd 2007, 01:52 PM
Jhevon
Quote:

Originally Posted by Mulya66
Any help/explanations would be grand.

"Evaluate the problem: log4 + 2log5"

i assume this is log to the base 10

log4 + 2log5
= log4 + log(5^2) ............................since n*log(x) = log(x^n)
= log4 + log25
= log(4*25) ....................................since log(x) + log(y) = log(xy)
= log100
= 2

Quote:

"Solve the equation: log5x-3log52=log53"
are these logs to the base 5?

if so,
log[5]x-3log[5]2=log[5]3
= log[5]x - log[5]2^3 = log[5]3 ...................since n*log(x) = log(x^n), i changed 3log[5]2 to log[5]2^3
= log[5]x - log[5]8 = log[5]3
= log[5]8x = log[5]3
now we can equate what's being logged.............since if logA = logB, then A = B, pretty intuitive i think
=> 8x = 3
=> x = 3/8

Quote:

Also, "Solve the equation: logx32= 5"
But as for this one, I tried and got as far as this:
x^5 = 32
i suppose this is log to the base x, you need to type these questions clearer, include the base in [] and let us know that's what it means

log[x]32 = 5
=> x^5 = 32 ...................................since if log[a]b = c, then a^c = b ...this is a fundamental law of logarithms, in fact, its the definition of a log
=> x = 32^(1/5)
=> x = 2

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if you have any questions, or if i misinterpreted any of the problems (since you did not identify the bases properly), don't hesitate to say so
• Apr 3rd 2007, 02:09 PM
Mulya66
Quote:

Originally Posted by Jhevon
if you have any questions, or if i misinterpreted any of the problems (since you did not identify the bases properly), don't hesitate to say so

You understood perfectly fine, thank you so much for your help and also letting me know to ask questions on the forum more clearly.
Thanks again!