Question

$\log 2=0.30103$, $\log 3=0.4771213\log 7=0.8450980$ then solve :
$\left(i\right)$ $2I^x=2^{2x+1}and2I^x=5^x$
$\left(ii\right)$ $3^{1+x}=7^{\frac{x}{2}}$

Answer 1

1. (a) $2I^x=2^{2x+1}$
$I^x=\frac{2^{2x+1}}{2}$
on solving,
$I^x=2^{2x+1-1}$
$I^x=4^x$
therefore, on compairing R.H.S with L.H.S
$I=4$
(b). $2I^x=5^x$
on taking log on both side,
$\log 2+x\log I=x\log 5$
$\log 2=x(\log 5-\log I)$
$0.30103=x(\log 10-\log 2-\log I)$
$\frac{0.30103}{x}=1-0.30103-\log I$
$\log I=0.6987-\frac{0.30130}{x}$

Therefore,
$I={10}^{0.6987-\frac{0.3087}{x}}$
(2). $3^{1+x}=7^{\frac{x}{2}}$
taking log on both sides,
$(1+x)\log 3=\frac{x}{2}\log 7$
on putting values,
$(1+x)0.4771213=\frac{x}{2}0.84509880$
$(1+x)0.4771213=\left(x\right)0.4225494$
$\frac{1+x}{x}=0.8856225869$
$\frac{1}{x}=-0.114377413$
therefore,
$x=-8.74298494$