+0  
 
0
133
1
avatar

1) Find \(\log_{5 \sqrt{5}}(625) \)

 

2) If \(\log_{10}(x) = 3 + \log_{10}(y)\) then find \(\dfrac{x}{y} \)

3) Let a and b be positive real numbers so that \( \log_b(x^2) = 10\) Find \(\log_{\sqrt[3]{b}} \left( \frac{1}{x} \right) \)

4)Find \(\frac{a}{b}\) when \(2\log{(a -2b)} = \log(a) + \log(b)\)

 Apr 6, 2019
 #1
avatar+106539 
0

(1)    log 5 √5    (625)   =    log 5 √5  (5)^4  

Using the change of base rule we can evaluate this in the following manner :

 

log 5^4             4 log 5              4 log 5                    4

______   =   ___________  =  __________  =    _____     = 4 (2/3)  =   8 / 3

log 5 √5            log 5^(3/2)        (3/2)  log 5            (3/2)

 

 

(2)    log x  = 3 + log y                subtract  log y  from both sides   

 

log x - log y  =  3

 

log (x / y) =  3

 

Note that we can write 3  as     log 1000

 

So

 

log (x / y)  =  log 1000

 

So   ....  (x/y)  = 1000 

 

 

(3)  log b x^2  = 10     

 

2 logb x  = 10            divide through by 2

 

log b x  = 5

 

This says that     b^5  =  x

 

So we have that

 

log ∛b  (1/ x)    =   log ∛b (1/b^5)  =   log b^(1/3) (b)^(-5)  =  (-5) log  b^(1/3)(b)  = -5 (3)  =  -15

 

 

(4)   2log(a - 2b)  = log a + log b

 

log(a - 2b)^2   =  log (ab)        

 

This implies that

 

(a - 2b)^2  =  ab          simplify

 

a^2 - 4ab + 4b^2  = ab

 

a^2 - 5ab + 4b^2  = 0        factor as

 

(a - 4b) ( a - b)  =  0 

 

The first term gives us what we need.....setting it to 0, we get that

 

a - 4b  =  0

 

a =  4b

 

a / b  =  4

 

 

 

cool cool cool

 Apr 6, 2019

33 Online Users

avatar
avatar
avatar