# Log question

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• Jun 22nd 2011, 05:09 AM
IBstudent
Log question
Hi,
This is a log problem that I can't solve..... I am kinda new to logs so any details (name of rule used....etc) would be very much appreciated. :-)

2(5^x+1) = 1 + 3/5^x

• Jun 22nd 2011, 05:17 AM
emakarov
Re: Log question
Replace 5^x with y and solve the equation for y. Then solve 5^x = y for the found value of y.
• Jun 22nd 2011, 05:22 AM
IBstudent
Re: Log question
Thanks... but can you please show me how to get a value for x, I constructed the equation you told me to.... but what do I do now???
• Jun 22nd 2011, 05:23 AM
Quacky
Re: Log question
Quote:

Originally Posted by IBstudent
Thanks... but can you please show me how to get a value for x, I constructed the equation you told me to.... but what do I do now???

Post some working if you wish to be assisted further.
• Jun 22nd 2011, 05:31 AM
IBstudent
Re: Log question
Quote:

Originally Posted by Quacky
Post some working if you wish to be assisted further.

fine....
I am sure it is 100% wrong, but since you insist.....
2(5^x+1) = 1 + 3/y
2(5^x+1) = y + 3/y
2y (5^x+1) = y+3
y( 5^x+1) = 3
5^2x+1 = 3 (I have a feeling I messed up bigtime, but I'll continue for your sake...)
(2x+1) log5 = log3
2x(log5) + log5 = log3
(stuck)

Thanks
• Jun 22nd 2011, 05:37 AM
emakarov
Re: Log question
I recommend replacing both occurrences of 5^x with y in the original equation and solving the result for y.
• Jun 22nd 2011, 05:38 AM
Quacky
Re: Log question
Quote:

Originally Posted by IBstudent
fine....
I am sure it is 100% wrong, but since you insist.....
2(5^x+1) = 1 + 3/y
2(5^x+1) = y + 3/y
2y (5^x+1) = y+3
y( 5^x+1) = 3
5^2x+1 = 3 (I have a feeling I messed up bigtime, but I'll continue for your sake...)
(2x+1) log5 = log3
2x(log5) + log5 = log3
(stuck)

Thanks

As emakarov said, remember that you let $5^x=y$? You can do that here too:
$2(5^x+1) = 1 + \frac{3}{y}$
$2(y+1)=1+\frac{3}{y}$
• Jun 22nd 2011, 05:48 AM
IBstudent
Re: Log question
Quote:

Originally Posted by Quacky
As emakarov said, remember that you let $5^x=y$? You can do that here too:
$2(5^x+1) = 1 + \frac{3}{y}$
$2(y+1)=1+\frac{3}{y}$

ok so:
2(y+1) = 1 + 3/y
2y+2 = 1+ 3/y
2y+1 = 3/y
cross multiplying
2y^2+y=3
2y^2+y-3=0
solving it using -b+-sqrt(b^2-4ac)/2a gives either 1 or -1.5. keeping in mind that this is 5^x. x could be 0 or.... undefined...
unfortunately, this is not the answer (not even close for that matter) :-(

Thanks
• Jun 22nd 2011, 06:00 AM
emakarov
Re: Log question
Your solution is correct: x = 0 is the only solution of this equation. You can check that it is a solution by substituting 0 for x in the original equation.
• Jun 22nd 2011, 06:11 AM
IBstudent
Re: Log question
sustituting gives 10=4
the mark scheme says that x is equivalent to -1 + log5(3) (by 5 I mean the base)
how is this possible?
• Jun 22nd 2011, 06:18 AM
emakarov
Re: Log question
Quote:

2(5^x+1) = 1 + 3/5^x
Left-hand side: 2(5^0 + 1) = 2(1 + 1) = 2 * 2 = 4.
Right-hand side: 1 + 3/5^0 = 1 + 3/1 = 1 + 3 = 4.
• Jun 22nd 2011, 06:21 AM
IBstudent
Re: Log question
oh I see thanks......
I guess the mark scheme is wrong and we are right!
• Jun 22nd 2011, 06:36 AM
emakarov
Re: Log question
On the other hand, the mark scheme is right if the equation is

$\displaystyle2\cdot5^{x+1} = 1 + \frac{3}{5^x}$,

which in ASCII can be represented as

2(5^(x+1)) = 1 + 3/5^x.

Then

$2\cdot 5^{x+1}=2\cdot5\cdot 5^x=10\cdot5^x$.

Replacing $5^x$ by y again and solving for y gives $y = 3/5$. So, $5^x=3/5$, or $5^{x+1}=3$, from where $x+1=\log_53$. Alternatively, $5^x=3/5$ gives

$x = \log_5(3/5) = \log_53-\log_55 = \log_53-1$.
• Jun 22nd 2011, 07:37 AM
IBstudent
Re: Log question
Last question....
so I know that x is -0.317394....... but, how did you know that this value+1 is equal to log5(3). It that a log rule? please tell me your method.
• Jun 22nd 2011, 07:54 AM
emakarov
Re: Log question
I did not claim above that $\log_53\approx −0.317394$, though this is true. If you don't have a calculator that computes logarithms to an arbitrary base, you can use the following equality.

$\log_ab\log_bc=\log_ac$ (1)

In particular,

$\log_53=\frac{\log_e3}{\log_e5}=\frac{\ln3}{\ln5}$. (2)

The mnemonic for (1) is transitivity: (a,b) and (b,c) gives you (a,c). For (2), the argument of ln() in the nominator becomes the argument of log() and the argument from the denominator becomes the base, which is located in a subscript below the log argument, just like the denominator is below the nominator.

Also, see Wikipedia.

Starting from $5^x=3/5$, you could also take ln() of both sides to get $\ln(5^x)=x\ln 5=\ln(3/5)=\ln3-\ln5$, from where $(x+1)\ln5=\ln3$ and $x=-1+\frac{\ln3}{\ln5}$.

Edit: I am not sure I addressed the question you asked. If not, feel free to rephrase it.
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