# Trouble

• May 7th 2012, 08:23 PM
bigtumbler
Trouble
Solve the equation

In10^x=Log e^x +2

Now I got

xIn10 = xLog e +2
x 2.302585093 = x 0.434294481 + 2

x = 0.434294481 +2/2.302585093

so x = 1.0572066

But imputting this into the original equation is incorrect. Where did I go wrong? Help would be much appreciated thank you.

Simon
• May 7th 2012, 08:44 PM
Prove It
Re: Trouble
Quote:

Originally Posted by bigtumbler
Solve the equation

In10^x=Log e^x +2

Now I got

xIn10 = xLog e +2
x 2.302585093 = x 0.434294481 + 2

x = 0.434294481 +2/2.302585093

so x = 1.0572066

But imputting this into the original equation is incorrect. Where did I go wrong? Help would be much appreciated thank you.

Simon

This is almost unreadable. What base is the logarithm on the right hand side? If it's base e, why not just use ln like on the LHS? Assuming it is...

\displaystyle \displaystyle \begin{align*} \ln{\left(10^x\right)} &= \ln{\left(e^x\right)} + 2 \\ \ln{\left(10^x\right)} - \ln{\left(e^x\right)} &= 2 \\ \ln{\left(\frac{10^x}{e^x}\right)} &= 2 \\ \ln{\left[\left(\frac{10}{e}\right)^x\right]} &= 2 \\ x\ln{\left(\frac{10}{e}\right)} &= 2 \\ x &= \frac{2}{\ln{\left(\frac{10}{e}\right)}} \\ x &= \frac{2}{\ln{10} - \ln{e}} \\ x &= \frac{2}{\ln{10} - 1} \end{align*}
• May 7th 2012, 08:56 PM
bigtumbler
Re: Trouble
This is how the equation was written to me. I am not sure why you used (In) on the RHS sinc In(e)=1 and log(e^1) = 0.434 approx.

Simon
• May 7th 2012, 09:29 PM
princeps
Re: Trouble
$\displaystyle \ln 10^x=\log_{10}e^x+2$

$\displaystyle x\ln 10=x\log_{10}e+2$

$\displaystyle x \cdot \left(\ln 10-\frac{1}{\ln 10}\right)=2$

Hence :

$\displaystyle x=\frac{2\ln 10}{\ln^2 10-1}$
• May 8th 2012, 12:07 AM
bigtumbler
Re: Trouble
Thank you very much :). I didn't know log10e = 1/In10. How come you used a minus instead of In10/(1/In10) a division sign? Sorry that my equations are not that clear. Unsure how to format it otherwise.
• May 8th 2012, 12:28 AM
princeps
Re: Trouble
Quote:

Originally Posted by bigtumbler
Thank you very much :). I didn't know log10e = 1/In10. How come you used a minus instead of In10/(1/In10) a division sign? Sorry that my equations are not that clear. Unsure how to format it otherwise.

$\displaystyle x\cdot a=x\cdot b+c \Rightarrow x\cdot a-x\cdot b=c\Rightarrow x \cdot(a-b)=c$
• May 8th 2012, 12:48 AM
bigtumbler
Re: Trouble
Ok so if it was for example 6x=4x +2 you would minus it (6x-4x=2) ? I always thought you had to divide because 6x=12 you divide.
• May 9th 2012, 12:33 PM
HallsofIvy
Re: Trouble
Do you not understand why you do those things? Your objective is to solve for x- that is, get x alone on one side. Since, in the second case, the only reason x is NOT alone is because it is multiplied by 6, it makes sense to divide by 6 to get 6x/6= 12/6. But that is NOT the case in "6x= 4x+ 2". You could divide both sides by 6- you would get x= (4/6)x+ (2/6)= (2/3)x+ (1/3). But because you still have x on both sides of the equation, that does not help a lot- and it has those annoying fractions. Instead, it is better to get rid of that "4x" on the right side. And because it is added you get rid of 4x by subtracting it from both sides:
6x- 4x= 4x+ 2- 4x or 2x= 2. Now what do you do to find x?

Also, going back to the original problem, you really need to distinguish between "I" (large letter "i") and "l" (small letter "L"). The symbol for the natural logarithm is "ln", not "In".