Question

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

Answer 1

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

Answer 2

ok ?

Answer 3

still don't get it

Answer 4

will get

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

Answer 5

or this is equal (t-1)^2

Answer 6

so than t-1=0

t=1

Answer 7

than logx 5=1

Answer 8

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

Answer 9

ok ?

Answer 10

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

\[\log_x5 + \frac{1}{\log_5x} = 2\]

\[(\log_x5)(\log_5x) + 1 = 2\log_5x\]

\[1 + 1 = 2\log_5x\]

\[\log_5x = 1\]
\[5^1 = x\]
\[x = 5\]

Answer 11

N.B

\[\log_ab \times \log_ba = 1\]

Answer 12

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

Answer 13

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

Answer 14

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