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

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

ok ?

still don't get it

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

or this is equal (t-1)^2

so than t-1=0 t=1

than logx 5=1

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

ok ?

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

N.B \[\log_ab \times \log_ba = 1\]

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

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

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

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