Mathematics
OpenStudy (anonymous):

(log base x argument 5 ) + ( 1 over log base 5 argument x) =2 algebraically solve for x

jhonyy9 (jhonyy9):

note logx 5 =t than will get t +1/t =2 and solve this for t after what you rewrite what will get for t make this values =logx 5 and calcule than the x

jhonyy9 (jhonyy9):

ok ?

OpenStudy (anonymous):

still don't get it

jhonyy9 (jhonyy9):

will get t^2 -2t +1 =0 solve this

jhonyy9 (jhonyy9):

or this is equal (t-1)^2

jhonyy9 (jhonyy9):

so than t-1=0 t=1

jhonyy9 (jhonyy9):

than logx 5=1

jhonyy9 (jhonyy9):

but you know that loga a =1 so than logx 5 =logx x from what result that x=5

jhonyy9 (jhonyy9):

ok ?

OpenStudy (anonymous):

(log base x argument 5 ) + ( 1 over log base 5 argument x) =2 algebraically solve for x $\log_x5 + \frac{1}{\log_5x} = 2$ $(\log_x5)(\log_5x) + 1 = 2\log_5x$ $1 + 1 = 2\log_5x$ $\log_5x = 1$ $5^1 = x$ $x = 5$

OpenStudy (anonymous):

N.B $\log_ab \times \log_ba = 1$

OpenStudy (turingtest):

...which can be derived from the change of base formula

OpenStudy (anonymous):

is that the same as this? $\log_ab \times \log_ba = \log_b{a^{\log_ab}} = \log_b{b} = 1$

OpenStudy (turingtest):

I was thinking$\log_ax=\frac{\log_bx}{\log_ba}$let x=b