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IB log help please!I
- Thread starter alexj987
- Start date
D
Deleted member 4993
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Hi,
I have been stuck on this problem for a while now with no luck:
Solve exactly for x: log(base 4)x^3 + log(base 2)sqrtx = 8
The answer given in the back of the book is 16 if that helps! I just need to be set in the right direction thank you!
Convert all the logs to base 2 ... and tell us what you get....
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Hello, alexj987!
We have: .\(\displaystyle \log_4(x^3) + \log_2\left(x^{\frac{1}{2}}\right) \:=\:8 \quad\Rightarrow\quad 3\log_4(x) + \frac{1}{2}\log_2(x) \:=\:8\)
Base-change: .\(\displaystyle 3\frac{\log_2(x)}{\log_2(4)} + \frac{1}{2}\log_2(x) \:=\:8 \quad\Rightarrow\quad 3\frac{\log_2(x)}{2} + \frac{\log_2(x)}{2} \:=\:8 \)
. . . . . . . . . . .\(\displaystyle 2\log_2(x) \:=\:8 \quad\Rightarrow\quad \log_2(x) \:=\:4\)
Therefore: . . .\(\displaystyle x \:=\:2^4 \quad\Rightarrow\quad x \:=\:16\)
\(\displaystyle \text{Solve for }x\!:\;\;\log_4(x^3) + \log_2(\sqrt{x}) \:=\: 8\)
We have: .\(\displaystyle \log_4(x^3) + \log_2\left(x^{\frac{1}{2}}\right) \:=\:8 \quad\Rightarrow\quad 3\log_4(x) + \frac{1}{2}\log_2(x) \:=\:8\)
Base-change: .\(\displaystyle 3\frac{\log_2(x)}{\log_2(4)} + \frac{1}{2}\log_2(x) \:=\:8 \quad\Rightarrow\quad 3\frac{\log_2(x)}{2} + \frac{\log_2(x)}{2} \:=\:8 \)
. . . . . . . . . . .\(\displaystyle 2\log_2(x) \:=\:8 \quad\Rightarrow\quad \log_2(x) \:=\:4\)
Therefore: . . .\(\displaystyle x \:=\:2^4 \quad\Rightarrow\quad x \:=\:16\)
HallsofIvy
Elite Member
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- Jan 27, 2012
- Messages
- 7,760
Did you learn anything from it? For example, you wrote, in your second post, "\(\displaystyle \frac{log_2(x)}{log_2(4)}\)" where soroban just wrote "\(\displaystyle \frac{1}{2}log_2(x)\)". Do you understand why \(\displaystyle log_2(4)= 2\)? If not then you really need to review the basic definitions of logarithms.