{"id":44497,"date":"2025-06-30T16:56:43","date_gmt":"2025-06-30T16:56:43","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=44497"},"modified":"2025-06-30T16:56:45","modified_gmt":"2025-06-30T16:56:45","slug":"solve-the-logarithm-equations-for-x","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/solve-the-logarithm-equations-for-x\/","title":{"rendered":"\u00a0Solve the logarithm equations for x."},"content":{"rendered":"\n
\u00a0Solve the logarithm equations for x.<\/pre>\n\n\n\n
\"\"<\/figure>\n\n\n\n
The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n
Here are the solutions to the logarithm equations.<\/p>\n\n\n\n
a) log\u2084(x\u00b2 – 6x) = 2<\/strong><\/p>\n\n\n\n
To solve this equation, we first convert it from logarithmic form to its equivalent exponential form. The relationship is log\u2090(b) = c is the same as a\u1d9c = b. Applying this to the given equation, the base is 4, the exponent is 2, and the argument is x\u00b2 – 6x.<\/p>\n\n\n\n
This gives us:
4\u00b2 = x\u00b2 – 6x
16 = x\u00b2 – 6x<\/p>\n\n\n\n
Now, we rearrange the equation into a standard quadratic form (ax\u00b2 + bx + c = 0) by subtracting 16 from both sides:
x\u00b2 – 6x – 16 = 0<\/p>\n\n\n\n
We can solve this quadratic equation by factoring. We look for two numbers that multiply to -16 and add to -6. These numbers are -8 and 2.
(x – 8)(x + 2) = 0<\/p>\n\n\n\n
This yields two possible solutions: x = 8 and x = -2.<\/p>\n\n\n\n
Finally, we must check both solutions in the original equation because the argument of a logarithm must be positive.
For x = 8: x\u00b2 – 6x = (8)\u00b2 – 6(8) = 64 – 48 = 16. Since 16 > 0, this solution is valid.
For x = -2: x\u00b2 – 6x = (-2)\u00b2 – 6(-2) = 4 + 12 = 16. Since 16 > 0, this solution is also valid.<\/p>\n\n\n\n
The solutions for (a) are x = 8 and x = -2.<\/strong><\/p>\n\n\n\n
b) log\u2082(x) + log\u2082(x – 4) = 5<\/strong><\/p>\n\n\n\n
This equation involves the sum of two logarithms with the same base. We use the product rule for logarithms, which states log\u2090(M) + log\u2090(N) = log\u2090(MN), to combine the terms on the left side.<\/p>\n\n\n\n
log\u2082(x(x – 4)) = 5
log\u2082(x\u00b2 – 4x) = 5<\/p>\n\n\n\n
Next, we convert this logarithmic equation to its exponential form:
2\u2075 = x\u00b2 – 4x
32 = x\u00b2 – 4x<\/p>\n\n\n\n
Rearrange into a standard quadratic equation:
x\u00b2 – 4x – 32 = 0<\/p>\n\n\n\n
Factor the quadratic equation. We need two numbers that multiply to -32 and add to -4. These are -8 and 4.
(x – 8)(x + 4) = 0<\/p>\n\n\n\n
The possible solutions are x = 8 and x = -4. We must check these against the original equation’s domain. The arguments of both logarithms, x and (x – 4), must be positive.
For x = 8: x = 8 (which is > 0) and x – 4 = 8 – 4 = 4 (which is > 0). This solution is valid.
For x = -4: x = -4 (which is not > 0). This is an extraneous solution and must be discarded.<\/p>\n\n\n\n
The solution for (b) is x = 8.<\/strong><\/p>\n\n\n\n
c) log\u2083(6 – x) – log\u2083(x + 4) = 2<\/strong><\/p>\n\n\n\n
We start by applying the quotient rule for logarithms, log\u2090(M) – log\u2090(N) = log\u2090(M\/N), to combine the terms.<\/p>\n\n\n\n
log\u2083((6 – x) \/ (x + 4)) = 2<\/p>\n\n\n\n
Convert to exponential form:
3\u00b2 = (6 – x) \/ (x + 4)
9 = (6 – x) \/ (x + 4)<\/p>\n\n\n\n
Now, solve for x. Multiply both sides by (x + 4):
9(x + 4) = 6 – x
9x + 36 = 6 – x<\/p>\n\n\n\n
Isolate the x term:
10x = -30
x = -3<\/p>\n\n\n\n
Finally, check if the solution is valid by ensuring the original arguments, (6 – x) and (x + 4), are positive.
For x = -3: 6 – x = 6 – (-3) = 9 (which is > 0) and x + 4 = -3 + 4 = 1 (which is > 0). The solution is valid.<\/p>\n\n\n\n
The solution for (c) is x = -3.<\/strong><\/p>\n\n\n\n
\"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"
\u00a0Solve the logarithm equations for x. The Correct Answer and Explanation is: Here are the solutions to the logarithm equations. a) log\u2084(x\u00b2 – 6x) = 2 To solve this equation, we first convert it from logarithmic form to its equivalent exponential form. The relationship is log\u2090(b) = c is the same as a\u1d9c = b. […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-44497","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/44497","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/comments?post=44497"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/44497\/revisions"}],"predecessor-version":[{"id":44506,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/44497\/revisions\/44506"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=44497"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=44497"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=44497"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}