Question

How to solve this.
5^x-2(5^-x)+1=0

Answer 1

use (ln)

Answer 2

\[5^{x}-2(5^{-x})+1=0\]
is this the equation

Answer 3

yes. then it would be:
\[[5^x-2(\frac{ 1 }{ 5^x })+1=0][5^x]\]

right?

Answer 4

why there is 5^x in right side

Answer 5

to eliminate the denominator

Answer 6

\[5^{x}-\frac{ 2 }{5^{x} }+1=0\]

Answer 7

\[5^{x}\left[ 5^{x} -\frac{ 2 }{ 5^{x}}+1\right]=5^{x}(0)\]
\[5^{2x}-2+5^{x}=0\]
\[\ln (5^{2x})-\ln (2)+\ln )5^{5}=\ln (0)\]
\[2x \ln (5)-\ln(2)+xln(5)=\ln(0)\]
\[x(2\ln(5)+\ln(5)=\ln(0)+\ln(2)\]
\[x=\frac{ \ln(0)+\ln(2) }{ 2\ln(5)+\ln(5) }\]

Answer 8

@blackbird02 try this change of variable:
\[5^{x}=y\]
where y is the new variable, substitute it in your equation, and you get a quadratic equation in y, the solve it for y

Answer 9

ok. thanks guys. i get it now