# What is x if log2x = log5 + log(x - 24/5) ?

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

First, we'll impose the constraints of existence of logarithms:

x - 24/5>0

x > 24/5

2x>0

x>0

The common interval of admissible values for x is (24/5 ,+inf.).

Now,we'll could solve the equation using the property of quotient:

log 2x = log 5 - log (x - 24/5)

log 2x  = log [5/(x - 24/5)]

Because the bases of logarithms are matching, we'll apply the one to one property:

2x = 5/(x - 24/5)

We'll cross multiply;

2x(x - 24/5) = 5

We'll remove the brackets:

2x^2 - 48x/5 - 5 = 0

We'll multiply by 5:

10x^2 - 48x - 25 = 0
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}