Evaluating and finding x; Logarithms

**yety124** · Jul 15th 2011, 02:27 PM

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?

thank you for your time

**mr fantastic** · Jul 15th 2011, 02:58 PM

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?

thank you for your time

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

**Soroban** · Jul 15th 2011, 05:46 PM

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]:

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

**yety124** · Jul 16th 2011, 03:29 AM

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?

**mr fantastic** · Jul 16th 2011, 02:10 PM

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.

**bjhopper** · Jul 16th 2011, 06:20 PM

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

**mr fantastic** · Jul 17th 2011, 12:17 AM

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.

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