# Solve for x log 4 x > 3 - log 2 x

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log 4 (x )> 3 - log 2 (x)

We know that:

log a b = log c b/ log c a

==> log 4 (x) = log 2 (x)/log 2 (4)

==> log 2 (x) /log 2 (4) > 3- log 2 (x)

But log 2 (4) = 2

==> log 2 (x) / 2 > 3 - log 2 (x)

Multiply by 2:

==> log 2 (x) > 6 - 2log 2 (x)

==> add 2 log 2 (x) to both sides:

==> log 2 (x) + 2 log 2 (x) > 6

==> 3log 2 (x) > 6

Divide by 3:

==> log 2 (x) > 2

but 2 = log 2 (4)

==> log 2 (x) > log 2 (4)

But log is an increasing function,

==> x > 4

==> x belongs to (4, inf)

Before solving the inequality, we'll impose the constraint of existence of logarithms, namely x>0.

We notice that the logarithms don't have the same base, so we'll create matching bases:

First, we'll express log 4 x = log 2 x / log 2 4

But log 2 4 = log 2 (2^2) = 2*log 2 2 = 2*1 = 2

log 4 (x) = log 2 (x) / 2

We'll move log 2 x to the left side and the inequality will become:

log 2 (x) + log 2 (x)/2 >3

We'll calculate the LCD and we'll add the terms:

3*log 2 (x) / 2>3

We'll divide by 3 both sides:

log 2 (x) / 2 > 1

We'll cross multiply:

log 2 (x) > 2

But 2 = log 2 (4)

log 2 (x) > log 2 (4)

The logarithms have the same base and is bigger than 1, so the function is increasing.

For log 2 (x) > log 2 (4) => x>4

**x belongs to the interval (4, +inf.).**

To solve for x. log 4x > 3- log2x.

Log ab = loga +logb; log a/b = loga-logb; log a^m = m loga.

So the given equation becomes:

log4+logx > 3- (log2+logx)

log4 +logx > 3 -log2 -logx.

logx+logx > 3 -log2 - log4

2logx > 3 - (log2+log4) = 3 -log4 = log10^3 - log4

2logx> log(10^3/8)

logx^2> log(10^3/8) . Take antilog.

x^2> 10^3/8

x> sqrt(10^3/8) = sqrt(125)

x > 5sqrt5.