+45 What is the real value of \(x\) in the equation \(log_224-log_23 = log_5x\)?
+36915 Switch the L side to log base 5
log5 24 / log 2 - log5 3/ log2
log5 (24/3) = log5 x
x = 24/3 = 8
+36915 Alan pointed out to me an error in this answer ....corrected below:
log5 24 / log5 2 - log5 3/ log5 2
= 3 = log5 x
5^3 = x = 125 (as UNTS subsequently showed below) Thanx , Alan!
+171 $ \log _2\left(24\right)-\log _2\left(3\right)=\log _5\left(x\right)\quad \overset{\Large{\text{if} \log_a(b)=c \: \therefore \: b=a^c }}{===============\Rightarrow }\quad \:5^{\log _2\left(24\right)-\log _2\left(3\right)}=x $
$5^{\log _2\left(\frac{24}{3}\right)}=x$
$5^{\log _2\left(8\right)}=x$
$5^{\log _2\left(2^3\right)}=x$
$5^{3}=x$
$\boxed{125=x}$
