Solving Logarithmic Equations: 2 Problems

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Homework Statement

I have 2 problems ,which are:
1) Solve :log ₇ (x ³+27) + log ₇ (x+3)
2) Solve for x:log₅ ( 5 ^1/x+ 125) log ₅ 6+1+1/2x

Homework Equations

The Attempt at a Solution

The first one tried like this :
log ₇ { (x ³+27)/(x+3) }= 2
or 7 ² = (x ³+27)/(x+3)
But I don't know how to proceed beyond it please help!
in second one
log₅{( 5 ^1/x+ 125) /6}=1+1/2x
but I can't proceed here too.
I need an urgent help!
Thank You!

 

[eumyang]

Please check the original problems as I see typos abound. For instance, in the first problem,

log ₇ (x ³+27) + log ₇ (x+3)

...I see no right hand side. Also, if what you have above is correct, then you're next step...

log ₇ { (x ³+27)/(x+3) }= 2

...is already wrong. (Is is supposed to be divide?)

Finally, if the last step above is supposed to have a divide, then review factoring special products -- specifically, the sum of two cubes.


69

 

[Mentallic]

For the first one, you could multiply through by x+3 to give

49(x+3)=x^3+27

and then solve that cubic by simplifying and turning it into the form ax^3+bx^2+cx+d=0
but cubics are hard to solve, an easier way would be to realize that x^3+27 is a sum of two cubes, mainly x^3+3^3=(x+3)(x^2-3x+9) which means you can cancel the x+3 from the numerator and denominator, giving you only a quadratic to solve, which is much simpler. Remember to look back at the original question to figure out what the domain of x is.

For the second, do the same thing as you did above, turn the log into powers such that

\frac{5^{\frac{1}{x}}+125}{6}=5^{1+\frac{1}{2x}}

Now simplify the right hand side. You should be able to factorize out 5^1/x and then solve.

 

[The Bob]

Homework Statement

I have 2 problems ,which are:
1) Solve :log ₇ (x ³+27) + log ₇ (x+3)
2) Solve for x:log₅ ( 5 ^1/x+ 125) log ₅ 6+1+1/2x

I agree with eumyang. Is the question to simply the above expressions? Also, with the second problem, are you missing any brackets/parenthesis at all?

The Bob

 

[HallsofIvy]

At least in the second, 1/2" specifically says "solve for x". There are no equations to be solved!

 

[Mentallic]

At least in the second, 1/2" specifically says "solve for x". There are no equations to be solved!

2) Solve for x:log₅ ( 5 ^1/x+ 125) log ₅ 6+1+1/2x

...

log₅{( 5 ^1/x+ 125) /6}=1+1/2x

It's an honest mistake to mix up + and = considering they're the same key on the keyboard, but he did have lots of typos so it could have been re-interpreted as anything.

And for the first equation, let's just assume he started to solve the correct problem, i.e. f(x)=2.

 

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Thanks for the help but I figured it out!
Thanks everyone!