FIND OUT THE VALUE OF "X" FROM THE EQUATION GIVEN BELOW: -
x^[log a to the base 2] + a^[log x to the base 2] = 2a^2
I tried taking Log to the bast 10 to both sides, but it didn't work out. Any idea how to do it????
FIND OUT THE VALUE OF "X" FROM THE EQUATION GIVEN BELOW: -
x^[log a to the base 2] + a^[log x to the base 2] = 2a^2
I tried taking Log to the bast 10 to both sides, but it didn't work out. Any idea how to do it????
Hello arijit2005We have to solve for $\displaystyle x$:$\displaystyle x^{\log_2a}+a^{\log_2x}=2a^2$If in doubt, get rid of logs as soon as you can - they're nasty things! So let $\displaystyle b = \log_2a$. Then $\displaystyle a=2^b$. So the equation becomes:
$\displaystyle x^b + \big(2^b\big)^{\log_2x} = 2\big(2^b\big)^2$Grandad
$\displaystyle \Rightarrow x^b + 2^{b\log_2x} = 2\cdot2^{2b}$
$\displaystyle \Rightarrow x^b + 2^{\log_2(x^b)} = 2\cdot2^{2b}$
$\displaystyle \Rightarrow x^b + x^b = 2\cdot(2^2)^b$
$\displaystyle \Rightarrow 2x^b = 2\cdot4^b$
$\displaystyle \Rightarrow x = 4$
Hi everyone,
Another approch:
If you use the logarithmic identity that I had mentioned in my previous post; $\displaystyle \huge{x^{log_{a}y}=y^{log_{a}x}}$
$\displaystyle x^{\log_2a}+a^{\log_2x}=2a^2$
$\displaystyle \Rightarrow{a^{\log_2x}+a^{\log_2x}=2a^2}$
$\displaystyle \Rightarrow{2a^{\log_2x}=2a^2}$
$\displaystyle \Rightarrow{a^{\log_2x}=a^2}$
$\displaystyle \Rightarrow{\log_2x=2}$
$\displaystyle \Rightarrow{x=4}$
Hope this will help you.
Hey... Thanks a lot man!!! No.. I haven't been introduced to this new formula $\displaystyle \huge{x^{log_{a}y}=y^{log_{b}x}}$
By the way, is it $\displaystyle \huge{x^{log_{a}y}=y^{log_{b}x}}$ or $\displaystyle \huge{x^{log_{a}y}=y^{log_{a}x}}$???
Will base "a" change?? You actually wrote $\displaystyle \huge{y^{log_{b}x}}$
Are you sure the $\displaystyle a$ becomes $\displaystyle b$???
Hello arijit2005Yes, it should be$\displaystyle x^{\log_ay}=y^{\log_ax}$You can prove this using the same substitution that I used (so that you can get rid of those nasty logs!):
Let $\displaystyle b = \log_ay$ and $\displaystyle c = \log_ax$Grandad
Then $\displaystyle a^b = y$ and $\displaystyle a^c = x$
So $\displaystyle x^{\log_ay} = x^b$$\displaystyle =(a^c)^b$
$\displaystyle = (a^b)^c$
$\displaystyle = y^c$
$\displaystyle =y^{\log_ax}$