You are not logged in.

- Topics: Active | Unanswered

Pages: **1**

**pman****Member**- Registered: 2005-11-03
- Posts: 10

hi can someone explain in detail how to solve this kind of problem please:

```
1) 7*4^x = 5*9^x
2)log3 3 + log9 1
```

you should use logarithms to solve it, thx!

Offline

**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 16,065

1) xlog4 + log7 = xlog9+ log5

xlog4 - xlog9 = log5 - log7

xlog9 - xlog4 = log7 - log5

x(log9 - log4) = log7 - log5

x = (log7 - log5)/(log9 - log4)

The second question isn't clear

Character is who you are when no one is looking.

Offline

**ryos****Member**- Registered: 2005-08-04
- Posts: 394

I think that, on the second question, he means "log base 3 of 3 + log base 9 of 1".

If so, it's easy. 3¹ = 3, and 9^0 = 1. So, 1 + 0 = 1.

El que pega primero pega dos veces.

Offline

**pman****Member**- Registered: 2005-11-03
- Posts: 10

thx alot guys

Offline

Pages: **1**