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

• nae99
In summary: Yes, if you also knew that the Quadratic Formula says that if you have two negative numbers, the sum of them is also negative.
nae99

## Homework Statement

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

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

## The Attempt at a Solution

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

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

what does that mean

nae99 said:

## Homework Statement

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

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

## The Attempt at a Solution

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

nae99 said:
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...

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

nae99 said:
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...

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

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

x^2 - 6x + 8= 2187

is that good

Looks good. Now solve for x.

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.

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

nae99 said:
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...

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

nae99 said:
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?

wen one that i am using is log 10

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

how do i substitue it

nae99 said:
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...

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

nae99 said:
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?

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

nae99 said:
the value of B=6
nae99 said:
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?

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

Last edited by a moderator:

nae99 said:
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?

x = 5 - (-5)
x = 10

nae99 said:
...
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 .

oh i see ok. so it should be;

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

nae99 said:
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.

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

ok thanks very much

No,

What is (-6)2 ?

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

How were you taking the square root of negative numbers?

nae99 said:
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

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