questions are on attachment!!
#6. Condense the expression to the logarithmic of a single quanity.
[2log 5 (x+4)+7log 5 (x+7)]-1/2log 5 x
Hello Brooke:
Use the properties of logs.
log(a)+log(b)=log(ab)...[1]
log(a)-log(b)=log(a/b)...[2]
Rewrite as:
$\displaystyle log_{5}(x+4)^{2}+log_{5}(x+7)^{7}-log_{5}x^{\frac{1}{2}}$
Using [1]
$\displaystyle log_{5}((x+4)^{2}(x+7)^{7})-log_{5}x^{\frac{1}{2}}$
Using [2]
$\displaystyle log_{5}(\frac{(x+4)^{2}(x+7)^{7}}{\sqrt{x}})$
Hello, Brooke!
1. Sketch the graph of the function: $\displaystyle f(x) \:= \:3 + e^x$
You're expected to be familiar with the graph of $\displaystyle y\:=\:e^x$.
This is the same graph translated up 3 units.
2. Evaluate the expression without using a calculator: $\displaystyle \log_{128}2$
Let: $\displaystyle \log_{128} \:=\;x$
Rewrite in exponential form: .$\displaystyle 128^x\:=\:2$
Get the same base on both sides: .$\displaystyle (2^7)^x\:=\:2\quad\Rightarrow\quad 2^{7x}\:=\:2^1$
Since the bases are equal, the exponents are equal: .$\displaystyle 7x\:=\:1\quad\Rightarrow\quad x\,=\,\frac{1}{7}$
Therefore: .$\displaystyle \log_{128}2\;=\;\frac{1}{7}$
3. Identify the logarithmic equation written in exponential form: $\displaystyle \log_{243}81\:=\:\frac{4}{5}$
Not sure what "identify" means here . . .
In exponential form, the expression is: .$\displaystyle 243^{\frac{4}{5}}\:=\:81$
4. Evaluate using the change-of-base formula: $\displaystyle \log_9517$
We have: .$\displaystyle \log_9517\:=\:\frac{\ln517}{\ln9}\:=\:2.843606857. ..$
5. Which is the logarithm rewritten as a ratio of natural logarithms?
. . . $\displaystyle A\!:\;\ln\left(\frac{3}{x}\right)\qquad B\!:\;\frac{\ln 3}{\ln x}\qquad C\!:$ $\displaystyle \;\ln x - \ln 3\qquad D\!:\;\frac{\ln x}{\ln 3}$
Is there a typo? . . . Both $\displaystyle B$ and $\displaystyle D$ are ratios of natural logs.
6. Condense the expression to the logarithm of a single quantity:
. . . $\displaystyle \left[2\log_5(x+4) + 7\log_5(x + 7)\right] - \frac{1}{2}\log_5x$
$\displaystyle \left[2\cdot\log_5(x+4) + 7\cdot\log_5(x+7)\right] - \frac{1}{2}\cdot\log_5(x)$
. . $\displaystyle = \;\left[\log_5(x+4)^2 + \log_5(x+7)^7\right] - \log_5\!\left(x^{\frac{1}{2}}\right)$
. . $\displaystyle = \;\log_5\!\left[(x+4)^2\cdot(x+7)^7\right] - \log_5\!\left(x^{\frac{1}{2}}\right)$
. . $\displaystyle = \;\log_5\!\left[\frac{(x+4)^2(x+7)^7}{x^{\frac{1}{2}}}\right] $