Having trouble getting started on a few of these.

1. 5^(x-10) = 125^x

2. (3/4) log x ^(3/4) + 9 = 0

thanks again!

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- May 10th 2009, 08:50 PMHapaSolve the following equation
Having trouble getting started on a few of these.

1. 5^(x-10) = 125^x

2. (3/4) log x ^(3/4) + 9 = 0

thanks again! - May 11th 2009, 09:03 AMinzaghina
Hello..

1) 125^x=5^(3x) => x-10=3x => x=-5

2) 3/4 * log x ^(3/4) + 9 = 0 => 3/4*3/4 * log x = -9 => 1/16 * log x = -1 => log x = -16

if log = ln => x=e^(-16) - May 11th 2009, 09:13 AMSoroban
Hello, Hapa!

Quote:

Equate exponents: .

Quote:

We have: .

. . . . . .

Edit: Too slow ... again! - May 11th 2009, 09:43 AMinzaghina
soroban, how do you do to write the equations so beautiful ? :)

- May 11th 2009, 04:11 PMlmasud
he used mathtype to write it like that

and here is another way to solve:

5^(x-10)= 125^x

5^(x-10)= 5^(3x)

Take "log" of both sides:

(x-10)log5= (3x)log5

dividing both sides by log5:

x-10= 3x

x= -5