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# I'm stuck on my log hw and i need help, URGENT

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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+0 Answers

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(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

Apr 6, 2019