Logarighms issues

[justan4cat]

The directions state: Express as a sum, difference, and product of logarighms, without using exponents.

log[sub:3v9h13vp]b[/sub:3v9h13vp] 4throot(x[sup:3v9h13vp]5[/sup:3v9h13vp]b[sup:3v9h13vp]9[/sup:3v9h13vp] / y[sup:3v9h13vp]7[/sup:3v9h13vp]z[sup:3v9h13vp]10[/sup:3v9h13vp])

It's a multiple choice question, and I can't seem to come up with any of the answers.

 

[BigGlenntheHeavy]

\(\displaystyle log_b\bigg[\frac{x^5b^9}{y^7z^{10}}\bigg]^{\frac{1}{4}}\)

\(\displaystyle = \ \frac{log_b\bigg[\frac{x^5b^9}{y^7z^{10}}\bigg]}{4}\)

\(\displaystyle = \ \frac{log_b(x^5b^9)-log_b(y^7z^{10})}{4}\)

\(\displaystyle = \ \frac{[log_b(x^5)+log_b(b^9)-(log_b(y^7)+log_b(z^{10})]}{4}\)

\(\displaystyle = \ \frac{[5log_b(x)+9log_b(b)-7log_b(y)-10log_b(z)]}{4}\)

\(\displaystyle = \ \frac{9+5log_b(x)-7log_b(y)-10log_b(z)}{4}\)

 

[soroban]

Hello, justan4cat!

Another approach . . .

Express as a sum, difference, and product of logarithms, without using exponents.

. . \(\displaystyle \log_b \sqrt[4]{ \frac{x^5b^9}{y^7z^{10}}}\)

\(\displaystyle \text{We have: }\;\log_b\!\left(\frac{x^5b^9}{y^7z^{10}}\right)^{\frac{1}{4}}\)

. . . . . \(\displaystyle =\;\log_b\!\left[\frac{(x^5)^{\frac{1}{4}}(b^9)^{\frac{1}{4}}}{(y^7)^{\frac{1}{4}}(z^{10})^{\frac{1}{4}}} \right]\)

. . . . . \(\displaystyle =\;\log_b\!\left[\frac{x^{\frac{5}{2}}\,b^{\frac{9}{4}}}{y^{\frac{7}{4}}\,z^{\frac{5}{2}}} \right]\)

. . . . . \(\displaystyle =\;\log_b\!\left(x^{\frac{5}{4}}b^{\frac{9}{4}}\right) - \log_b\!\left(y^{\frac{7}{4}}z^{\frac{5}{2}}\right)\)

. . . . . \(\displaystyle =\; \left[\log_b\!\left(x^{\frac{5}{4}}\right) + \log_b\!\left(b^{\frac{9}{4}}\right)\right] - \left[\log_b\!\left(y^{\frac{7}{4}}\right) + \log_b\!\left(z^{\frac{5}{2}}\right)\right]\)

. . . . . \(\displaystyle =\; \log_b\!\left(x^{\frac{5}{4}}\right) + \log_b\!\left(b^{\frac{9}{4}}\right) - \log_b\!\left(y^{\frac{7}{4}}\right) - \log_b\!\left(z^{\frac{5}{2}}\right)\)

. . . . . \(\displaystyle =\;\tfrac{5}{4}\log_b(x) + \tfrac{9}{4}\underbrace{\log_b(b)}_{\text{This is 1}} - \tfrac{7}{4}\log_b(y) - \tfrac{5}{2}\log_b(z)\)

. . . . .\(\displaystyle =\;\tfrac{5}{4}\log_b(x) + \tfrac{9}{4} - \tfrac{7}{4}\log_b(y) - \tfrac{5}{2}\log_b(z)\)

 

[justan4cat]

Soroban... Thank you so very much! I understand that now! However, now that I see how you worked that, it confuses me on the next problem on my homework. It looks almost the same, and when I work it like you did, I'm not getting one of the choice questions. Here's what I have:
Express as a single logarithm, and, if possible, simplify.

log[sub:3utpvstv]a[/sub:3utpvstv]y + 1/7 log[sub:3utpvstv]a[/sub:3utpvstv]z

using: log[sub:3utpvstv]a[/sub:3utpvstv]M + log[sub:3utpvstv]a[/sub:3utpvstv]N=log[sub:3utpvstv]a[/sub:3utpvstv](MN)
log[sub:3utpvstv]a[/sub:3utpvstv]y + 1/7 log[sub:3utpvstv]a[/sub:3utpvstv]z
log[sub:3utpvstv]a[/sub:3utpvstv]y + log[sub:3utpvstv]a[/sub:3utpvstv](z[sup:3utpvstv]1/7[/sup:3utpvstv])
log[sub:3utpvstv]a[/sub:3utpvstv]y(z[sup:3utpvstv]1/7[/sup:3utpvstv])

but the choices have a 4 in them, and I don't understand where that is coming from. The choices are:

log[sub:3utpvstv]a[/sub:3utpvstv]y[sup:3utpvstv]4[/sup:3utpvstv]/z[sup:3utpvstv]1/7[/sup:3utpvstv]
log[sub:3utpvstv]a[/sub:3utpvstv](y[sup:3utpvstv]4[/sup:3utpvstv]z[sup:3utpvstv]1/7[/sup:3utpvstv])
log[sub:3utpvstv]a[/sub:3utpvstv](y[sup:3utpvstv]1/7[/sup:3utpvstv]z[sup:3utpvstv]4[/sup:3utpvstv])
log[sub:3utpvstv]a[/sub:3utpvstv](yz)[sup:3utpvstv]4/7[/sup:3utpvstv]

I have to be missing something.