- #1
nae99
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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
Bassalisk said:This is trivial stuff in my opinion. Domain x>4. (x-2)*(x-4)=3^7. well rest is...
nae99 said: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
nae99 said:what does that mean
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
nae99 said:iam lose, don't know where to go frm here
nae99 said:x^2 - 6x + 8= 2187
x^2 - 6x +8 - 2187 =0
x^2 - 6x - 2179 =0
x = - 6 [itex]\pm[/itex] [itex]\sqrt{}[/itex] 6^2 - 4*1*2179[itex]/[/itex] 2*1
x = - 6 ± [itex]\sqrt{}[/itex] 36 - 8716 [itex]/[/itex] 2*1
x = - 6 ± [itex]\sqrt{}[/itex] -8680 [itex]/[/itex] 2
x = -6 ± 93.167[itex]/[/itex] 2
x = -6 + 93.167[itex]/[/itex] 2
x = 87.167 [itex]/[/itex] 2
X = 4.436
how is that so far
nae99 said:i don't know how to plug it into the equation
nae99 said:that link does not work and what error is in the eqaution
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
nae99 said:the value of B=6
nae99 said:and C is 2179
x^2 - 6x - 2179 =0
nae99 said:i am confuse...so are you saying that i should change the minus to plus
nae99 said:...
x^2 - 6x - 2179 =0
x = - 6 [itex]\pm[/itex] [itex]\sqrt{}[/itex] 6^2 - 4*1*2179[itex]/[/itex] 2*1
Use enough parentheses to make your expressions say what you mean.nae99 said:oh i see ok. so it should be;
x = -(-6) [itex]\pm[/itex] [itex]\sqrt{}[/itex] -6^2 - 4*1*-2179 [itex]/[/itex] 2*1
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).nae99 said:i don't know how to plug it into the equation