Help With Logarithm

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PheelGoodInc

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Hi, thanks for having me on the site. I'm reviewing for a test and am stuck on two logarithms.

logarithms.JPG

The two in question are highlighted aboce.

Number 16, I keep trying to raise 2x^2+3? I'm getting nowhere with that and google hasn't helped much. (it's much harder to search google for logarithm problems due to the format of the text).

Number 19, I have tried cancelling the logs since they have the same bse.

I get 2x-8=20
Simplified to 12. The answer sheet says it's 10.

Any help is greatly appreciated. Thank you.
 
PheelGoodInc said:
Hi, thanks for having me on the site. I'm reviewing for a test and am stuck on two logarithms.

View attachment 5329

The two in question are highlighted aboce.

Number 16, I keep trying to raise 2x^2+3? I'm getting nowhere with that and google hasn't helped much. (it's much harder to search google for logarithm problems due to the format of the text).

Number 19, I have tried cancelling the logs since they have the same bse.

I get 2x-8=20
Simplified to 12. The answer sheet says it's 10.

Any help is greatly appreciated. Thank you.
Click to expand...

(18)

\(\displaystyle \displaystyle{2^{x^2}\cdot 8 \ = \ 4^{2x}}\) →

\(\displaystyle \displaystyle{\ 2^{x^2}\cdot 2^3 \ = \ 4^{2x}}\)→

\(\displaystyle \displaystyle{\ 2^{x^2 + 3} \ = \ (2^2)^{2x}}\)→

\(\displaystyle \displaystyle{\ 2^{x^2 + 3} \ = \ 2^{4x}}\)→

Continue....

(19)

\(\displaystyle \displaystyle{2Log_{11}(x-4) \ = \ Log_{11}2 + Log_{11}18}\)→

\(\displaystyle \displaystyle{Log_{11}[(x-4)^2] \ = \ Log_{11}(2*18)}\)→

\(\displaystyle \displaystyle{Log_{11}[(x-4)^2] \ = \ Log_{11}(36)}\)→ Now you can eliminate "Log"s.
 
Last edited by a moderator:
PheelGoodInc said:
Hi, thanks for having me on the site. I'm reviewing for a test and am stuck on two logarithms.

View attachment 5329

The two in question are highlighted aboce.

Number 16, I keep trying to raise 2x^2+3? I'm getting nowhere with that and google hasn't helped much. (it's much harder to search google for logarithm problems due to the format of the text).

Number 19, I have tried cancelling the logs since they have the same bse.

I get 2x-8=20
Simplified to 12. The answer sheet says it's 10.

Any help is greatly appreciated. Thank you.
Click to expand...

Just a different way for (18)
\(\displaystyle 2^{x^2}\, 8\, =\, 4^{2x}\, \Rightarrow\, ln_2(2^{x^2}\, 8)\, =\, ln_2(4^{2x})\, \Rightarrow\)
\(\displaystyle ln(2^{x^2})\, +\, ln_2(8)\, =\, {2x}\, ln_2(4)\Rightarrow\, ...\)
 
Ishuda said:
Just a different way for (18)
\(\displaystyle 2^{x^2}\, 8\, =\, 4^{2x}\, \Rightarrow\, ln_2(2^{x^2}\, 8)\, =\, ln_2(4^{2x})\, \Rightarrow\)
\(\displaystyle ln(2^{x^2})\, +\, ln_2(8)\, =\, {2x}\, ln_2(4)\Rightarrow\, ...\)
Click to expand...

"ln" looks like the abbreviation for natural log. And there is a "sub 2" missing from left side of the bottom equation.


The use of \(\displaystyle \ log_2 \ \) makes it clear.

OP, this is actually problem # 16.


\(\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2^{x^2}*8 \ = \ 4^{2x} \ \implies\)

\(\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \log_2(2^{x^2}*8) \ = \ \log_2(4^{2x}) \ \implies\)

\(\displaystyle \log_2(2^{x^2}) + \log_2(8) \ = \ \log_2(4^{2x}) \ \implies\)

\(\displaystyle \log_2(2^{x^2}) + \log_2(2^3) \ = \ \log_2[(2^2)^{2x}] \ \implies\)

\(\displaystyle \log_2(2^{x^2}) + \log_2(2^3) \ = \ \log_2(2^{4x}) \ \implies\)

\(\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x^2 + 3 \ = \ 4x \ \implies \ \ \) ?
 
lookagain said:
"ln" looks like the abbreviation for natural log. And there is a "sub 2" missing from left side of the bottom equation.


The use of \(\displaystyle \ log_2 \ \) makes it clear.

OP, this is actually problem # 16.


\(\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2^{x^2}*8 \ = \ 4^{2x} \ \implies\)

\(\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \log_2(2^{x^2}*8) \ = \ \log_2(4^{2x}) \ \implies\)

\(\displaystyle \log_2(2^{x^2}) + \log_2(8) \ = \ \log_2(4^{2x}) \ \implies\)

\(\displaystyle \log_2(2^{x^2}) + \log_2(2^3) \ = \ \log_2[(2^2)^{2x}] \ \implies\)

\(\displaystyle \log_2(2^{x^2}) + \log_2(2^3) \ = \ \log_2(2^{4x}) \ \implies\)

\(\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x^2 + 3 \ = \ 4x \ \implies \ \ \) ?
Click to expand...

Answer sheet has it as (1,3). How is that?
 
PheelGoodInc said:
Also, the answer sheet has 10 for number 19. I'm not even close.
Click to expand...

Use Subhotosh Khan's suggestion to eliminate the \(\displaystyle log_{11}\). That will give you a quadratic. One of the roots is "a false root introduced by squaring the (x-4)", so check both roots with the original equality.

For example, suppose I have
x - 4 = 6
then
(x - 4)2 = 36
If I use just the second equality, x can be either -2 or +10 and, by checking back with the original equation, we see than x=10 is the proper solution. Now, if the original equation had been
|x - 4| = 6
then both solutions would be correct.
 
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