Hard logarithm problem HELP!

by **Learningforcomp** » Fri Sep 11, 2020 3:06 pm

Greetings, I was learning logarithms when I found this problem(pls see attachment) and it ask me to simplify it, but i have not found yet an explanation on how to solve it.
Could you please provide tips or an explanation as to how to solve this problem(and the steps if possible) please?
The help is appreciated.
Thank you.

by **Baltuilhe** » Fri Sep 11, 2020 4:12 pm

Good afternoon!

Look for 'change of base'.
[tex]\log_b(a)=\frac{\log_c{a}}{\log_c{b}}[/tex]

So:
[tex]\log_{x_2}\,x_1\cdot\log_{x_3}\,x_2\cdot\ldots\cdot\log_{x_n}\,x_{n-1}=\frac{\log\,x_1}{\log\,x_2}\cdot\frac{\log\,x_2}{\log\,x_3}\cdot\ldots\cdot\frac{\log\,x_{n-1}}{\log\,x_n}=\frac{\log\,x_1}{\log\,x_n}=\log_{x_n}\,x_1[/tex]

by **Learningforcomp** » Sat Sep 12, 2020 8:17 am

Baltuilhe wrote:Good afternoon!

Look for 'change of base'.
<math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mstyle displaystyle="true" scriptlevel="0"><msub><mo><mi>log</mi><mo>⁡</mo></mo><mi>b</mi></msub><mo stretchy="false">(</mo><mi>a</mi><mo stretchy="false">)</mo><mo>=</mo><mfrac><mrow><msub><mo><mi>log</mi><mo>⁡</mo></mo><mi>c</mi></msub><mi>a</mi></mrow><mrow><msub><mo><mi>log</mi><mo>⁡</mo></mo><mi>c</mi></msub><mi>b</mi></mrow></mfrac></mstyle></mrow><annotation encoding="application/x-tex">\displaystyle \log_b(a)=\frac{\log_c{a}}{\log_c{b}}</annotation></semantics></math>logb(a)=logcblogca

So:
<math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mstyle displaystyle="true" scriptlevel="0"><msub><mo><mi>log</mi><mo>⁡</mo></mo><msub><mi>x</mi><mn>2</mn></msub></msub> <msub><mi>x</mi><mn>1</mn></msub><mo>⋅</mo><msub><mo><mi>log</mi><mo>⁡</mo></mo><msub><mi>x</mi><mn>3</mn></msub></msub> <msub><mi>x</mi><mn>2</mn></msub><mo>⋅</mo><mo>…</mo><mo>⋅</mo><msub><mo><mi>log</mi><mo>⁡</mo></mo><msub><mi>x</mi><mi>n</mi></msub></msub> <msub><mi>x</mi><mrow><mi>n</mi><mo>−</mo><mn>1</mn></mrow></msub><mo>=</mo><mfrac><mrow><mi>log</mi><mo>⁡</mo> <msub><mi>x</mi><mn>1</mn></msub></mrow><mrow><mi>log</mi><mo>⁡</mo> <msub><mi>x</mi><mn>2</mn></msub></mrow></mfrac><mo>⋅</mo><mfrac><mrow><mi>log</mi><mo>⁡</mo> <msub><mi>x</mi><mn>2</mn></msub></mrow><mrow><mi>log</mi><mo>⁡</mo> <msub><mi>x</mi><mn>3</mn></msub></mrow></mfrac><mo>⋅</mo><mo>…</mo><mo>⋅</mo><mfrac><mrow><mi>log</mi><mo>⁡</mo> <msub><mi>x</mi><mrow><mi>n</mi><mo>−</mo><mn>1</mn></mrow></msub></mrow><mrow><mi>log</mi><mo>⁡</mo> <msub><mi>x</mi><mi>n</mi></msub></mrow></mfrac><mo>=</mo><mfrac><mrow><mi>log</mi><mo>⁡</mo> <msub><mi>x</mi><mn>1</mn></msub></mrow><mrow><mi>log</mi><mo>⁡</mo> <msub><mi>x</mi><mi>n</mi></msub></mrow></mfrac><mo>=</mo><msub><mo><mi>log</mi><mo>⁡</mo></mo><msub><mi>x</mi><mi>n</mi></msub></msub> <msub><mi>x</mi><mn>1</mn></msub></mstyle></mrow><annotation encoding="application/x-tex">\displaystyle \log_{x_2}\,x_1\cdot\log_{x_3}\,x_2\cdot\ldots\cdot\log_{x_n}\,x_{n-1}=\frac{\log\,x_1}{\log\,x_2}\cdot\frac{\log\,x_2}{\log\,x_3}\cdot\ldots\cdot\frac{\log\,x_{n-1}}{\log\,x_n}=\frac{\log\,x_1}{\log\,x_n}=\log_{x_n}\,x_1</annotation></semantics></math>logx2x1⋅logx3x2⋅…⋅logxnxn−1=logx2logx1⋅logx3logx2⋅…⋅logxnlogxn−1=logxnlogx1=logxnx1

Thank you Baltuilhe. I was able to understand and find the answer, I deeply appreciate your help.

Hard logarithm problem HELP!

Hard logarithm problem HELP!

Re: Hard logarithm problem HELP!

Re: Hard logarithm problem HELP!

Who is online