# Solve for x log3 (x-2) + log3 (x-4) = 7

1. Jul 6, 2011

### nae99

1. The problem statement, all variables and given/known data
i have been trying to do this ques. 2 days now and still cant get the ans.
help plz

Solve for x
log3 (x-2) + log3 (x-4) = 7

2. Relevant equations

3. The attempt at a solution
log3 (x-2) (x-4) = 7

2. Jul 6, 2011

### Staff: Mentor

Re: logarithms

3. Jul 6, 2011

### nae99

Re: logarithms

what does that mean

4. Jul 6, 2011

### Staff: Mentor

Re: logarithms

His hint was to use the inverse function of the logarithm as your first step.

So if you were working with log base 2, you would raise both sides to the exponent power of 2....

$$log_2 (x) = 5$$

$$2^{log_2 (x)} = 2^5$$

...and simplify the lefthand side...

5. Jul 6, 2011

### nae99

Re: logarithms

i would then prolly do this

x^2-4x-2x-8 = 3^7

x^2- 6x- 8 = 3^7

am i on the right track

6. Jul 6, 2011

### Staff: Mentor

Re: logarithms

There's a sign error in your multiplication of (-2)*(-4), but other than that...

7. Jul 6, 2011

### nae99

Re: logarithms

iam lose, dont know where to go frm here

8. Jul 6, 2011

9. Jul 6, 2011

### nae99

Re: logarithms

x^2 - 6x + 8 = 3^7

x^2 - 6x + 8= 2187

is that good

10. Jul 6, 2011

### SammyS

Staff Emeritus
Re: logarithms

Looks good. Now solve for x.

11. Jul 6, 2011

### Staff: Mentor

Re: logarithms

And once you solve for x, use your scientific calculator to check your answer by plugging it back into your original problem statement in Post #1.

12. Jul 6, 2011

### nae99

Re: logarithms

x^2 - 6x + 8= 2187

x^2 - 6x +8 - 2187 =0

x^2 - 6x - 2179 =0

x = - 6 $\pm$ $\sqrt{}$ 6^2 - 4*1*2179$/$ 2*1

x = - 6 ± $\sqrt{}$ 36 - 8716 $/$ 2*1

x = - 6 ± $\sqrt{}$ -8680 $/$ 2

x = -6 ± 93.167$/$ 2

x = -6 + 93.167$/$ 2

x = 87.167 $/$ 2

X = 4.436

how is that so far

13. Jul 6, 2011

### Staff: Mentor

Re: logarithms

You tell us. Is the answer correct when you plug it into the original equation in Post #1?

EDIT -- I do think I see a sign error in the first line where you apply the Quadratic Formula, though...

14. Jul 6, 2011

### nae99

Re: logarithms

i dont know how to plug it into the equation

15. Jul 6, 2011

### Staff: Mentor

Re: logarithms

First, fix the sign error(s) in the Quadratic Equation application, and then, do you have a scientific calculator that lets you do log base n?

16. Jul 6, 2011

### Staff: Mentor

17. Jul 6, 2011

### nae99

Re: logarithms

wen one that i am using is log 10

18. Jul 6, 2011

### nae99

Re: logarithms

that link does not work and what error is in the eqaution

19. Jul 6, 2011

### nae99

Re: logarithms

how do i substitue it

20. Jul 6, 2011

### Staff: Mentor

Re: logarithms

The link works fine when I click on it. But in any case, all you have to do is Google logarithm calculator, and you get lots of useful links (inclucding that one).

You have two sign errors that I think I see in your application of the Quadratic Equation. Just be careful in handling negative signs, and re-check the line where you substitute A, B and C into the Quadratic Equation...