2loga x3 - log x2 - 3 log x
can you please explain to me how to solve this equation?
thank you so much <3
can you please explain to me how to solve this equation?
thank you so much <3
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Assume all logs to same base a.
log x^3 =3logx
logx^2 = 2logx
So 2logx^3 - logx^2 -3logx = 6logx - 2logx - 3logx = logx
Hope this helps.
log x^3 =3logx
logx^2 = 2logx
So 2logx^3 - logx^2 -3logx = 6logx - 2logx - 3logx = logx
Hope this helps.
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Are they all log base a?? And do you mean:
x3 = x³, x2 = x²
...if so, then just use the addition/subtraction rules:
log(a) + log(b) = log(a * b) and log(a) - log(b) = log(a / b)
furthermore: nlog(a) = log(a^n)
-->
2log_a(x³) = log_a((x³)²) = log_a(x⁶)
3log_a(x) = log_a(x³)
-->
2log_a(x³) - log_a(x²) - 3log_a(x)
-->
log_a(x⁶) - log_a(x²) - log_a(x³)
-->
log_a(x⁶ / (x² * x³)) = log_a(x⁶ / x⁵) = log_a(x)
x3 = x³, x2 = x²
...if so, then just use the addition/subtraction rules:
log(a) + log(b) = log(a * b) and log(a) - log(b) = log(a / b)
furthermore: nlog(a) = log(a^n)
-->
2log_a(x³) = log_a((x³)²) = log_a(x⁶)
3log_a(x) = log_a(x³)
-->
2log_a(x³) - log_a(x²) - 3log_a(x)
-->
log_a(x⁶) - log_a(x²) - log_a(x³)
-->
log_a(x⁶ / (x² * x³)) = log_a(x⁶ / x⁵) = log_a(x)
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trick: logx + logy = log(xy)
and, logx - logy = log(x/y)
2loga x3 - log x2 - 3 log x
log(a2 x6) - log(x2) - log(x3)
log(a^2 c^6) - log(x6)
log(a2)
2*log(a)
and, logx - logy = log(x/y)
2loga x3 - log x2 - 3 log x
log(a2 x6) - log(x2) - log(x3)
log(a^2 c^6) - log(x6)
log(a2)
2*log(a)
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2log_a x^3 - log_a x^2 - 3 log_a x =
log_a (x^3)^2 - log_a x^2 - log_a x^3 =
log_a (x^3)^2 / x^2*x^3 =
log_a (x^6 / x^5) =
log_a x
log_a (x^3)^2 - log_a x^2 - log_a x^3 =
log_a (x^3)^2 / x^2*x^3 =
log_a (x^6 / x^5) =
log_a x