The number of solution (s) of the equation lo | vedclass

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Question

$2^{x}=t$

$\log _{7}(t-1)+\log _{7}(x->)=1$

$\log [(t-1)(x-7)]=1$

$(x-1)(t-2)=7$

$k^{2}-7 t-x+7=7$

$t^{2}-8 t=0$

$t_{1}=0$ or $t=

Vedclass · Accepted Answer

$1$

Vedclass · Answer

$2^{x}=t$

$\log _{7}(t-1)+\log _{7}(x->)=1$

$\log [(t-1)(x-7)]=1$

$(x-1)(t-2)=7$

$k^{2}-7 t-x+7=7$

$t^{2}-8 t=0$

$t_{1}=0$ or $t=8$

$2^{x}=0 \quad$ or $\quad 2^{x}=8$

$2^{x}=2^{3}$

$\Rightarrow x=3$

The number of solution $(s)$ of the equation $log_7(2^x -1) + log_7(2^x -7) = 1$, is -

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