Did I solve these logarithms correctly?

[wrongnmbr]

1. 35log[base 6](x) = 18
My answer: x = 6^18/35

2. 6^(-0.2x) - 3 = 7
A bit stuck on this one. I get as far as 6^(-0.2x) = 9, but I'm a bit lost at finding the x?

3. 7log[base 5](x) - 3 = 15
My answer: x = 5^12/7

 

[lookagain]

1. 35log[base 6](x) = 18
My answer: x = 6^18/35 . . . . \(\displaystyle This \ \ is \ \ not \ \ correctly \ \ typed. \ \ What \ \ you \ \ have \ \ typed \ \ is \ \ the \ \ same \ \ as\)


\(\displaystyle x = \frac{6^{18}}{35} \ \ \ \\) \(\displaystyle \ \\)Put grouping symbols around the exponent, such as "x = 6^(18/35)."


2. 6^(-0.2x) - 3 = 7
A bit stuck on this one. I get as far as 6^(-0.2x) = 9, but I'm a bit lost at finding the x? . . . . \(\displaystyle It \ \ should \ \ be \ \ \ \\)\(\displaystyle \ \ \ \ \ \\) "6^(-0.2x) = 10."

Then \(\displaystyle \log_{10}6^{-0.2x} = \log_{10}10 \longrightarrow -0.2x\log_{10}6 = 1 \ \ \ \ (Optional \ \ log \ \ base)\)

\(\displaystyle Continue \ \ solving \ \ this...\)



3. 7log[base 5](x) - 3 = 15
My answer: x = 5^12/7

\(\displaystyle 3. \ \ \ \ No,\ \ it \ \ would \ \ become \ \\)

\(\displaystyle 7\log_{5}(x) = 18 \longrightarrow \ \ \log_{5} = \frac{18}{7} \longrightarrow\ \or \ \ x = 5^{\frac{18}{7}} \ \ or \ \\) x = 5^(18/7).

\(\displaystyle \ \ \ \ Again, \ \ you \ \ need \ \ grouping \ \ symbols. \ \\) \(\displaystyle \ \\) \(\displaystyle Note \ \ the \ \ first \ \ problem.\)