Math Help - Trouble

1. 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

2. Re: Trouble

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 \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*}

3. 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

4. Re: Trouble

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

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

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

Hence :

$x=\frac{2\ln 10}{\ln^2 10-1}$

5. 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.

6. Re: Trouble

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.
$x\cdot a=x\cdot b+c \Rightarrow x\cdot a-x\cdot b=c\Rightarrow x \cdot(a-b)=c$

7. 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.

8. 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".