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kakasveta [241]
3 years ago
9

What is the true solution to the equation below?

Mathematics
1 answer:
Naddik [55]3 years ago
5 0
Yeeee

assuming your equaiton is
2ln(e^{ln(2x)})-ln(e^{ln(10x)})=ln(30)


remember some nice log rules
log_a(b)=c translates to a^c=b
and
a^{log_a(b)}=b
and
xlog_c(b)=log_c(b^x)
and
ln(x)=log_e(x)
and
log(a)-log(b)=log(\frac{a}{b})
and
if log(a)=log(b) then a=b

so

we can simplify a bit of stuff here

the e^{ln(2x)} \space\ and \space\ the \space\ e^{ln(10x)} can be simplified to 2x \space\ and \space\ 10x

so we gots now

2ln(2x)-ln(10x)=ln(30)
ln((2x)^2)-ln(10x)=ln(30)
ln(4x^2)-ln(10x)=ln(30)
ln(\frac{4x^2}{10x})=ln(30)
same base so
\frac{4x^2}{10x}=30
\frac{2x}{5}=30
times both sides by 5
2x=150
divide both sides by 2
x=75
answer is x=75
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