Question

Logarithm problem?

2log(base 9) + 1 = 2 log (base x) 3

My answer doesn't match the correct answer (which is x = 1/9 or x = 3). Are my steps right?

2 log (base 9) x + 1 = 1/(log (base 9) x)
2 log (base 9) x + 1 log (base 9) 9 = 1/(log (base 9) x)

Answer 1

\[2 \log_9 x + 1 = 2 \log_x 3\]
\[2 \log_9 x + 1\log_99= \frac{1}{\log_9x}\]

Answer 2

Cross multiply. Make the right hand side. You'll get a quadratic in (log(base9)x)

Answer 3

\[2(\log_{9}x)^{2}+\log_{9} 9\times \log_{9}x=1 \]

Answer 4

easiest way for 1 is to use change of base on the right hand side
\[2\log _{x}3 = \log _{x} 3^2 = \log _{x}9\]

then \[\log _{x} 9 = \log _{9} 9/\log _{9}x = 1/\log _{9}x\]

Answer 5

I brought it over to make it 1/log (base 9) x. Was that wrong? D:

Answer 6

Oh, that's correct. Solve the quadratic I wrote. What do you get?

Answer 7

so the problem is now

\[\log _{9} x^2 + 1 = 1/\log _{9} x\]

\[(\log _{9}x^2)\times(\log _{9} x) + \log _{9} x = 1\]

take the exponent of every term

\[x^2 \times x + x = 9\]

so x^3 + x = 9

Answer 8

I got x=3 and 1/9, exactly what you wanted

Answer 9

Hey campbell, how did you 'take the exponent of every term'?

Answer 10

Hold up one sec, let me try it out on pen and paper. :))

Answer 11

Ah, ne'er mind. :)) I got it. Thank you everyone!

Answer 12

I don't understand @campbell_st sol'n, though. D:

Answer 13

I used change of base... all to base 9.... then its the raised to the exponent of 9

Answer 14

I think one sol'n would do. :)) Thank you very much, nonetheless!

Answer 15

(log9x2)×(log9x)+log9x=1
take the exponent of every term
x2×x+x=9
Can you really do that, Campbell?

Answer 16

my mistake I didn't recognise the equation that was reducible to a quadratic...

2u^2 + u - 1 = 0 where u = log9 (x)

Answer 17

yep... they are all base 9 terms

Answer 18

(log 10)*log(100)+(log 1000)=5(Base is 10)
but 10*100+1000=2000?

Answer 19

isn't the mistake log(10)x log(100) = log10^(log100)