Question

7^log x =98-x^log7 calculate the value of x

Answer 1

Answer:

$x=e^2$

Step-by-step explanation:

Concept used:

$N=a^x$

Then, $log_aN=x$

$N=a^{log_aN}$

Given:

$7^{logx}=98-x^{log7}$

$7^{logx}+x^{log7}=98$

$(e^{log7})^{logx}+(e^{logx})^{log7}=98$

$e^{log7\:logx}+e^{logx\:log7}=98$

$e^{log7\:logx}+e^{log7\:logx}=98$

$2(e^{log7\:logx})=98$

$e^{log7\:logx}=49$

$(e^{log_e7})logx=49$

$7^{logx}=7^2$

This implies,

$logx=2$

$x=e^2$

Answer 2

Answer:

Value of x is e²

Step-by-step explanation:

Given : $7^{log\,x}=98-x^{log\,7}$

We use the following results to find value of x.

$(x^a)^b=x^{ab}$

$e^{log\,x}=x$

Consider,

$7^{log\,x}=98-x^{log\,7}$

$7^{log\,x}+x^{log\,7}=98$

$(e^{log\,7})^{log\,x}+(e^{log\,x})^{log\,7}=98$

$e^{log\,7\:log\,x}+e^{log\,x\:log\,7}=98$

$(e^{log\,7})^{log\,x}=49$

$(7)^{log\,x}=7^2$

$log\,x=2$

$e^{log\,x}=e^2$

$x=e^2$

Therefore, Value of x is e²