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

[nae99]

Homework Statement

i have been trying to do this ques. 2 days now and still can't get the ans.
help please

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

Homework Equations

The Attempt at a Solution

log3 (x-2) (x-4) = 7

 

[berkeman]

A reply from another thread:

This is trivial stuff in my opinion. Domain x>4. (x-2)*(x-4)=3^7. well rest is...

 

[nae99]

what does that mean

 

[berkeman]

Homework Statement

i have been trying to do this ques. 2 days now and still can't get the ans.
help please

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

Homework Equations

The Attempt at a Solution

log3 (x-2) (x-4) = 7

what does that mean

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...

 

[nae99]

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

 

[berkeman]

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

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

 

[nae99]

iam lose, don't know where to go frm here

 

[berkeman]

iam lose, don't know where to go frm here

After fixing the sign error, solve the quadratic equation for x using the Quadratic Formula:

http://en.wikipedia.org/wiki/Quadratic_equation

.

 

[nae99]

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

x^2 - 6x + 8= 2187

is that good

 

[SammyS]

Looks good. Now solve for x.

 

[berkeman]

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.

 

[nae99]

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

 

[berkeman]

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

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...

 

[nae99]

i don't know how to plug it into the equation

 

[berkeman]

i don't know how to plug it into the equation

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

 

[berkeman]

If not, there are lots of online calculators:

http://www.rapidtables.com/calc/math/Log_Calculator.htm

.

 

[nae99]

wen one that i am using is log 10

 

[nae99]

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

 

[nae99]

how do i substitue it

 

[berkeman]

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

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...

 

[nae99]

the mistake that I am seeing is here:
x = -6 ± 93.167/ 2

it should have been
x = -6± -93.167/ 2

but i keep looking where i make the mistake when i substitute A B C but i can't recognize it

 

[berkeman]

the mistake that I am seeing is here:
x = -6 ± 93.167/ 2

it should have been
x = -6± -93.167/ 2

but i keep looking where i make the mistake when i substitute A B C but i can't recognize it

What is the value of B? What is the Quadratic Formula in terms of A, B and C? What is the sign of C? Did it get propagated into the last term in the numerator inside the square root?

 

[nae99]

the value of B=6 , the formula http://en.wikipedia.org/wiki/Quadratic_equation

 

[nae99]

and the value of A is x which is = 1 and C is 2179

 

[berkeman]

the value of B=6

and C is 2179

No.

The equation that you ended up with was:

x^2 - 6x - 2179 =0

Look again at the definition of the Quadratic Formula. Are there minus signs in the definition?

 

[nae99]

i am confuse...so are you saying that i should change the minus to plus

 

[berkeman]

i am confuse...so are you saying that i should change the minus to plus

What do you think?

If you had an equation x = 5 - y, and I told you y = -5, what would x be?

 

[nae99]

x = 5 - (-5)
x = 10

 

[SammyS]

...
x^2 - 6x - 2179 =0

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

x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

a = 1, b = -6, c = -2179

So, you made a sign errors when plugging in both b and c .

 

[nae99]

oh i see ok. so it should be;

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

 

[nae99]

 

[SammyS]

oh i see ok. so it should be;

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

Use enough parentheses to make your expressions say what you mean.

If you mean, x = {-(-6) ± √[(-6)^2 - 4*1*(-2179) ] }/ (2*1),

then, yes, you are right.

 

[nae99]

and now it will be:
x = -(-6) \pm \sqrt{} -36 - 8716 / 2

ok thanks very much

 

[SammyS]

No,

What is (-6)² ?

What is -4*(1)*(-2179) ?

How were you taking the square root of negative numbers?

 

[Ray Vickson]

i don't know how to plug it into the equation

If log[a](b) denotes the base-a logarithm of b, then x = log[a](b) solves the equation a^x = b = (exp(ln(a))^x = exp(x*ln(a)), so x*ln(a) = ln(b). So, if you can compute ln then x = log[a](b) = ln(b)/ln(a). If you prefer to use base-10 logs, you would have, instead, log[a](b) = log[10](b)/log[10](a).

RGV

 

[nae99]

(-6)^2 = -36

-4*(1)*(-2179) = 8716

so should it be √ -36 + 8716 / 2

 

[nae99]

no,

what is (-6)² ?

What is -4*(1)*(-2179) ?

How were you taking the square root of negative numbers?

(-6)^2 = 36

-4*(1)*(-2179) = 8716

 

[BloodyFrozen]

Or you can use change of base. It might be a little tedious though...

 

[SammyS]

Or you can use change of base. It might be a little tedious though...

Have you looked at the previous posts in this thread? Change of base would appear to be a huge stretch if we are having this trouble with plugging into the quadratic formula.

 

[nae99]

Have you looked at the previous posts in this thread? Change of base would appear to be a huge stretch if we are having this trouble with plugging into the quadratic formula.

i don't have any how to plug it into anything, so I am clueless of what you u are talking about

 

[BloodyFrozen]

Have you looked at the previous posts in this thread? Change of base would appear to be a huge stretch if we are having this trouble with plugging into the quadratic formula.

Yeah, true...

(-6)^2 = -36
-4*(1)*(-2179) = 8716

so should it be √ -36 + 8716 / 2

(-6)^2 = (-1)^2*(6)^2 = 36
-6^2 = -1*(6)^2 = -36

P.S. Nae99, may I ask what kind of Pre-calc class (college, high school) you are taking?

 

[nae99]

Yeah, true...



(-6)^2 = (-1)^2*(6)^2 = 36
-6^2 = -1*(6)^2 = -36

P.S. Nae99, may I ask what kind of Pre-calc class (college, high school) you are taking?

college