Questions. (1 Viewer)

[Lemiixem]

Solve logx^2 + log8x = 3

The diagram shows the graph of a certain function y=f(x). Sketch the graph of f'(x).

The graph looks similar to a y=sinx graph except that its been pushed to the left a bit.

 

[Carrotsticks]

Solve logx^2 + log8x = 3

The diagram shows the graph of a certain function y=f(x). Sketch the graph of f'(x).

The graph looks similar to a y=sinx graph except that its been pushed to the left a bit.

1. What base is the logarithm in? I presume e?

2. A sine graph shifted to the left a bit, so did you mean a cosine graph?

 

[Lemiixem]

The logarithm is in base 10.

The graph looks like this.

 

[deswa1]

<a href="http://www.codecogs.com/eqnedit.php?latex=logx^2@plus;log8x=3\\ 2logx@plus;log8@plus;logx=3\\ 3logx=3-log8\\ logx=1-log8^{\frac{1}{3}} \\logx=log10-log2\\ logx=log5\\ x=5" target="_blank"><img src="http://latex.codecogs.com/gif.latex?logx^2+log8x=3\\ 2logx+log8+logx=3\\ 3logx=3-log8\\ logx=1-log8^{\frac{1}{3}} \\logx=log10-log2\\ logx=log5\\ x=5" title="logx^2+log8x=3\\ 2logx+log8+logx=3\\ 3logx=3-log8\\ logx=1-log8^{\frac{1}{3}} \\logx=log10-log2\\ logx=log5\\ x=5" /></a>

 

[Carrotsticks]

We have to of course utilise log properties:

 

[Carrotsticks]

For the graphing one, since your curve *sorta* looks like a negative cosine curve, our derivative curve is going to look like a positive sine curve, so draw one!

 

[Lemiixem]

For the second line from the bottom of your working, how did you go from:
1 - log 2 To log10 - log 2?

 

[Carrotsticks]

Because: