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Mathematics 8 Online
OpenStudy (anonymous):

I need help solving: solve terms in natural logarithms log4 x + log 11 x=1

zepdrix (zepdrix):

\[\large \log_4x+\log_{11}x=1\]So is this what the problem looks like? and they want us tooooooo convert to natural log to find x? :o

OpenStudy (anonymous):

yes

zepdrix (zepdrix):

We'll use the change of base formula,\[\large \log_ab=\frac{\log_cb}{\log_ca}\] Where c is any value we want. In this problem we want c=e. The log of base e is the natural log.

zepdrix (zepdrix):

So for the first term, applying this rule gives us,\[\large \log_4x= \frac{\log_e x}{\log_e 4}\]Which we can of course simply write as,\[\large \frac{\ln x}{\ln 4}\]

zepdrix (zepdrix):

So we'll get something similar for the next term, giving us an equation of,\[\large \frac{\ln x}{\ln4}+\frac{\ln x}{\ln11}=1\]

zepdrix (zepdrix):

So the wording at the start is a little confusing. Do they want us to solve for \(x\)? or can we stop when we get to \(\ln x\)?

OpenStudy (anonymous):

Solve for x and decimal approximation

zepdrix (zepdrix):

From our last step, we'll factor ln x out of each term on the left,\[\large \ln x\left(\frac{1}{\ln4}+\frac{1}{\ln11}\right)=1\] If we get a common denominator inside the brackets, we'll have,\[\large \ln x\left(\frac{\ln11+\ln4}{\ln4\cdot\ln11}\right)=1\]Multiplying both sides by the RECIPROCAL of what's in the brackets gives us,\[\large \ln x = \left(\frac{\ln4\cdot\ln11}{\ln4+\ln11}\right)\] From here, we can exponentiate both sides: Rewrite both sides with a base e,\[\huge e^{\ln x} = e^{\left(\frac{\ln4\cdot\ln11}{\ln4+\ln11}\right)}\]

zepdrix (zepdrix):

On the left, the exponential and logarithm are inverse operations of one another, so they essentially cancel out.\[\huge x = e^{\left(\frac{\ln4\cdot\ln11}{\ln4+\ln11}\right)}\]

zepdrix (zepdrix):

Then we just have to punch this into the calculator to solve for x. It's a bit tough to get through all the steps :( I know.

OpenStudy (anonymous):

i got 5.75 is that right?

zepdrix (zepdrix):

Hmmm no. Maybe having a little calculator trouble there.

OpenStudy (anonymous):

oh wait 7.80

OpenStudy (anonymous):

4.80 oops

zepdrix (zepdrix):

\[\large \text{e ^}((\ln(4)\ln(11))/(\ln(4)+\ln(11))\] It depends how you're putting it into the calculator. If you want to put it all in at once, it will look like this. It's a lot of brackets so it can be a bit confusing :)

zepdrix (zepdrix):

The number you're trying to get issssss, 2.40714491 So try to get that out of your calculator c:

OpenStudy (anonymous):

got it...I switched calc's .....cell phone calculator was doing it ...thank you!

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