Simplifying Logarithmic Function y=14log2(8x−56)16−12 by applying log properties

[NewToMath]

Simplifying Logarithmic Function y=14log2(8x−56)16−12 by applying log properties

Hello, here is the exact problem that I am struggling with:

Simplify the equation [FONT=MathJax_Math]y[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main]4[/FONT][FONT=MathJax_Main]log[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Main]8[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]56[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]16[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]12[/FONT]by applying the laws of logarithms. How are the original equation and the simplified equation related?

I have had no problems simplifying logarithmic expressions before when I am solving for an unknown variable.

The issue here appears to be that I don't recognize any of the laws for logarithms at work that I can reverse that wouldn't fundamentally change the function. The simplified version has to be equivalent to the original.

I have tried many different angles, such as transforming it to exponential form, but none of them have worked for me.

I put the question into WolframAlpha, but when I put the answer it gave me into graphing software, the simplified version and the original did not map onto each other.

Thank you for any help that you can provide. [FONT=MathJax_Math]y[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main]4[/FONT][FONT=MathJax_Main]log[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Main]8[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]56[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]16[/FONT]

 

[Ishuda]

Hello, here is the exact problem that I am struggling with:

Simplify the equation [FONT=MathJax_Math]y[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main]4[/FONT][FONT=MathJax_Main]log[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Main]8[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]56[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]16[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]12[/FONT]by applying the laws of logarithms. How are the original equation and the simplified equation related?

I have had no problems simplifying logarithmic expressions before when I am solving for an unknown variable.

The issue here appears to be that I don't recognize any of the laws for logarithms at work that I can reverse that wouldn't fundamentally change the function. The simplified version has to be equivalent to the original.

I have tried many different angles, such as transforming it to exponential form, but none of them have worked for me.

I put the question into WolframAlpha, but when I put the answer it gave me into graphing software, the simplified version and the original did not map onto each other.

Thank you for any help that you can provide. [FONT=MathJax_Math]y[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main]4[/FONT][FONT=MathJax_Main]log[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Main]8[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]56[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]16[/FONT]

Just what is the equation?
y = \(\displaystyle 14\, log_2[(8x-56)*16]\, -\, 12\)
or what? If that is the equation, what is log(a*b) and 16 = 2^?

 

[Majasaro]

So you want to simplify y=14log₂[(8x-56)16]-12 factor 8 in the brackets
y=14log₂[(x-7)16*8]-12
16*8=2⁴*2³=2³⁺⁴=2⁷
y=14log₂[(x-7)2^7]-12
Now applying log(ab)=log(a)+log(b)
y=14[log₂(x-7)+log₂2^7]-12
Using log(a^b)=b[log(a)], This is something important [log(a)]^b≠[log(a)]
y=14[log₂(x-7)+7log₂2]-12
logarithm of a Number same than his base is 1, This is log₂2=1
y=14[log₂(x-7)+7]-12
Now distribute the 14
y=14log₂(x-7)+98-14
and finally
y=14log₂(x-7)+84

 

[stapel]

I put the question into WolframAlpha, but when I put the answer it gave me into graphing software, the simplified version and the original did not map onto each other.

It probably couldn't understand your input...?

Simplify the equation y=14log2(8x−56)16−12 by applying the laws of logarithms. How are the original equation and the simplified equation related?

As another helper noted, "the equation" is difficult to interpret. Do you mean either of the following?

. . . . .\(\displaystyle \mbox{a. }\, y\, =\, 14\, \log_2\left((8x\, -\, 56)^{16}\right)\, -\, 12\)

. . . . .\(\displaystyle \mbox{b. }\, y\, =\, 14\, \left(\log_2(8x\, -\, 56)\right)^{16}\, -\, 12\)

. . . . .\(\displaystyle \mbox{c. }\, y\, =\, 14\, \log\big(2\left((8x\, -\, 56)^{16}\right)\big)\, -\, 12\)

Or something else?

When you reply, please include a clear listing of your thoughts and efforts so far. Thank you!