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Basic of Logarithms
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The number of solution $(s)$ of the equation $log_7(2^x -1) + log_7(2^x -7) = 1$, is -
A
$0$
B
$1$
C
$2$
D
$3$
Solution
$2^{x}=t$
$\log _{7}(t-1)+\log _{7}(x->)=1$
$\log [(t-1)(x-7)]=1$
$(x-1)(t-2)=7$
$k^{2}-7 t-x+7=7$
$t^{2}-8 t=0$
$t_{1}=0$ or $t=8$
$2^{x}=0 \quad$ or $\quad 2^{x}=8$
$2^{x}=2^{3}$
$\Rightarrow x=3$
Standard 11
Mathematics