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What is the real value of \(x\) in the equation \(log_224-log_23 = log_5x\)?

 Jul 3, 2021
 #1
avatar+37147 
0

Switch the L side to log base 5

 

log5  24 / log 2     -    log5 3/ log2  

log5 (24/3) = log5 x

 

x = 24/3   = 8

 Jul 3, 2021
 #3
avatar+37147 
0

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!  

ElectricPavlov  Jul 3, 2021
 #2
avatar+171 
+4

$ \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 $

 

$5^{\log _2\left(\frac{24}{3}\right)}=x$

 

$5^{\log _2\left(8\right)}=x$

 

$5^{\log _2\left(2^3\right)}=x$

 

$5^{3}=x$

 

$\boxed{125=x}$

 Jul 3, 2021

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