Hi Ppl. Please could you help me solve this log question..?

Question

Its the following:

log x to base 2 + 1=6log 2 to base x

Any help is appreciated. The answer should be 4 or an eighth. Thanks

Answer 1

Both 4 and 1/8 are correct answers and the only answers.

Solution:

The trick is to know this "conversion of log-base" formula:

log_a (y) = log_b (y) / log_b(a)

Then you can do this:

6 log_x (2) = log_x (2^6) = log_2 (2^6)/log_2(x) = 6/log_2(x)

Now write y = log_2(x). Substituting this into the original equation, you get

y + 1 = 6/y

This is a quadratic with two roots: y = 2 and -3.

The corresponding x is then equal to

x = 2^y; that is 2^2 = 4 or 2^{-3} = 1/8.

Both are correct.

Answer 2

well i got the ans as 1/8 or 4
log x to base 2 + 1 = 6 / (log x to base 2)

take log x to base 2 = y
so y + 1 = 6 / y
and y^2 + y - 6 = 0

so y = -3,2
so log x to base 2 = y =-3 or 2
so 2^-3 = 1/8 = x
or 2^2 = 4 = x
got ur ans
hope it helped!

Answer 3

log x (base 2) = 1/(log 2 (base x)) so...

log x (base 2) + 1 = 6/(log x (base 2))

Assume now all logs are base 2:

(log x)^2 + log x -6 = 0

say that A=log x

A^2 + A - 6 = 0 (quadratic)

A = 2 or A= -3

so A=2 and now log x (base 2) = 2, so x=4
or A= -3, and x= 1/8

Answer 4

logx base2+1= log64 basex
" " = log64 base2 / logx base2
Using substitution, let logx base2=a
hence, a+1= 6/a
a2 +a-6=0 solve this quadratic equation. a=-3,2
logx base2 =-3, 2 so x=4, 1/8

hope it helps.