# Thread: Evaluating and finding x; Logarithms

1. ## Evaluating and finding x; Logarithms

a.) so i was trying to evaluate this
(log5 4 x log2 10)/log25 sqrt10

so what i did was do change of base on the numerator than i got 4 (is doing this correct?) then after i got this well i got stuck on what to do next. help?

b.) one last i came across this problem log3 (x+2)=3logx 3
i got stuck on this as well, what would i do to find x?

2. ## Re: Evaluating and finding x; Logarithms

Originally Posted by yety124
[snip]

b.) one last i came across this problem log3 (x+2)=3logx 3
i got stuck on this as well, what would i do to find x?

The right hand side is $\log_x (27)$. Using the change of base rule, this can be written as $\frac{3}{\log_3 (x)}$ ....

3. ## Re: Evaluating and finding x; Logarithms

Hello, yety124!

$\text{(a) Simplify: }\:\frac{\log_54 \cdot \log_210}{\log_{25}\sqrt{10}}$

$\text{We have: }\:\frac{\log_5(4) \cdot \log_2(10)}{\log_{25}(10^{\frac{1}{2}})}$ .[1]

. . $\log_5(4) \:=\:\log_5(2^2) \:=\:2\log_5(2)$

. . $\log_2(10) \:=\:\frac{\log_5(10)}{\log_5(2)}$

. . $\log_{25}(10)^{\frac{1}{2}} \:=\:\frac{\log_5(10^{\frac{1}{2}})}{\log_5(25)} \:=\:\frac{\log_5(10^{\frac{1}{2}})}{\log_5(5^2)} \;=\;\frac{\frac{1}{2}\log_5(10)}{2\,\log_5(5)}} \:=\:\frac{\log_5(10)}{4}$

Substitute into [1]:

. . $\frac{2\,\rlap{//////}{\log_5(2)}\cdot\dfrac{\rlap{///////}\log_5(10)}{\rlap{//////}\log_5(2)}} {\dfrac{\rlap{///////}\log_5(10)}{4}} \;=\;8$

4. ## Re: Evaluating and finding x; Logarithms

ok i get the first one now thanks
but then the second one tried to solve it this is where i got to
log3 (x+2) * log3 (x)=3
what do i do after this?

5. ## Re: Evaluating and finding x; Logarithms

Originally Posted by yety124
ok i get the first one now thanks
but then the second one tried to solve it this is where i got to
log3 (x+2) * log3 (x)=3
what do i do after this?
At this point I'd re-check the question, and also check that there was not a typo in the question.

6. ## Re: Evaluating and finding x; Logarithms

HI yety124,
For the second one
logB3 (x+2) =3logBx (3)
log B3 (x+2) =log Bx (27)= logB3 (27)/logB3(x)
logB3 (x+2)* log B3 (x) = logB3 (27)
x^2 + 2x = 27

7. ## Re: Evaluating and finding x; Logarithms

Originally Posted by bjhopper
HI yety124,
For the second one
logB3 (x+2) =3logBx (3)
log B3 (x+2) =log Bx (27)= logB3 (27)/logB3(x)
logB3 (x+2)* log B3 (x) = logB3 (27)
x^2 + 2x = 27
The left hand side of the final line follows from an invalid log rule and is wrong. Products of logs do not simplify in the way this line suggests .....

If the * had been a +, then the last line would be correct.