# So far i have x+3 ln 10=7-x ln 5e, does the e cancle out when I log both sides to get 1/2=2x-4? Equation: 10^(x+3)=5e^7(-x)

So far i have $x+3\mathrm{ln}10=7-x\mathrm{ln}5e$ , does the e cancle out when I log both sides to get $\frac{1}{2}=2x-4$ ?
equation: ${10}^{x+3}=5{e}^{7-x}$
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Step 1
According to you by taking natural log on both sides of the above $e{q}^{n}$ you got
$x+3\mathrm{ln}10=7-x\mathrm{ln}5e$ but it is wrong you didn't applied the logarithm property correctly
On appliying log on both sides of $e{q}^{n}$ x we get
appx