Question

How to solve log equations?

`1. log _{x}125 = log _{8}x^3`
`2. log x^2 = log _{2x}25`
`3. log _{15}x = log _{5}3x`

Answer 1

Generally, it helps to reduce both sides to a common base.

Explanation:

There is no overall answer to this question. Some log equations are best solved by a range of methods. However, in general, it usually helps to reduce both sides to a common base.

To demonstrate this I will solve your question 1.

`log_x 125 = log_8 x^3`

Let's reduce both sides to `ln`.

Remember `log_a b = ln b/ln a`

`:. ln 125/ln x = ln(x^3)/ln8 = (3lnx)/ln8`

`3ln^2 x = ln125 xx ln8`

`3ln^2 x approx 10.0402`

`ln^2 x approx 3.34673`

`lnx approx +- sqrt3.34673`

`lnx approx +-1.82941`

`x approx e^(+1.82941) or e^(-1.82941)`

`x approx 6.2302 or 0.1605`

NB: This approach can be used on the other two problems. In Q2 you will need to assume a base for the LHS - I suggest 10, but it must be stated for the solution to be valid.

Answer 2

(1) `6.23`; (2) `x=5` or `0.1` and (3) `x=1/15`

Explanation:

(1) As `log_x125=log_8x^3` can be written as

`log_x5^3=log_8x^3` or `3log_x5=3log_8x`

or `log_x5=log_8x`

i.e. `log5/logx=logx/log8`

or `(logx)^2=log5log8`

= `0.6990xx0.9031=0.6312669`

i.e. `logx=sqrt0.6312669=0.7945`

and `x=10^0.7945=6.23`

(2) `logx^2=log_(2x)25`

i.e. `2logx=2log_(2x)5`

or `logx=log5/(log2x)=log5/(log2+logx)`

or `(logx)^2+log2logx-log5=0`

and `logx=(-log2+-sqrt((log2)^2+4*log5))/2`

= `(-0.3010+-sqrt((0.3010)^2+4*0.6990))/2`

= `(-0.301+-1.699)/2`

= `0.699` or `-1` i.e. `x=10^0.699` or `10^(-1)`

Hence. `x=5` or `0.1`

(3) `log_15x=log_5 3x`

i.e. `logx/log15=(log3+logx)/log5`

or `logxlog5=log15log3+log15logx`

or `logx(log15-log5)=-log15log3`

or `logxlog3=-log15log3`

i.e. `logx=-log15=log(1/15)`

Hence `x=1/15`