Solve \\( 5^x = 100x – 175 \\)<\/p>\n\n\n\n
Alright \u2014 let\u2019s solve \\( 5^x = 100x – 175 \\) from scratch using the Lambert W method systematically, without guessing solutions.<\/p>\n\n\n\n
—<\/p>\n\n\n\n
1. Rewrite the equation
We start with:
\\[
5^x = 100x – 175
\\]
\\[
e^{x \\ln 5} = 100x – 175
\\]<\/p>\n\n\n\n
—<\/p>\n\n\n\n
2. Eliminate the constant term in the linear part
We want the right-hand side to be of the form \\( \\alpha (x – \\beta) \\).
Factor 100:
\\[
100x – 175 = 100\\left(x – \\frac{175}{100}\\right) = 100\\left(x – \\frac{7}{4}\\right)
\\]
So the equation becomes:
\\[
e^{x \\ln 5} = 100\\left(x – \\frac{7}{4}\\right)
\\]<\/p>\n\n\n\n
—<\/p>\n\n\n\n
3. Shift the variable
Let \\( t = x – \\frac{7}{4} \\), so \\( x = t + \\frac{7}{4} \\).<\/p>\n\n\n\n
Substitute:
\\[
e^{\\left(t + \\frac{7}{4}\\right) \\ln 5} = 100 t
\\]
\\[
e^{t \\ln 5} \\cdot e^{\\frac{7}{4} \\ln 5} = 100 t
\\]
\\[
5^{7\/4} \\cdot e^{t \\ln 5} = 100 t
\\]
\\[
e^{t \\ln 5} = \\frac{100}{5^{7\/4}} \\, t
\\]<\/p>\n\n\n\n
Let \\( k = \\frac{100}{5^{7\/4}} \\), so:
\\[
e^{t \\ln 5} = k \\, t
\\]<\/p>\n\n\n\n
—<\/p>\n\n\n\n
4. Rearrange into Lambert W form
\\[
1 = k \\, t \\, e^{-t \\ln 5}
\\]
\\[
\\frac{1}{k} = t \\, e^{-t \\ln 5}
\\]
Multiply both sides by \\(-\\ln 5\\):
\\[
-\\frac{\\ln 5}{k} = (-t \\ln 5) \\, e^{-t \\ln 5}
\\]<\/p>\n\n\n\n
Let \\( u = -t \\ln 5 \\), then:
\\[
-\\frac{\\ln 5}{k} = u \\, e^{u}
\\]<\/p>\n\n\n\n
—<\/p>\n\n\n\n
5. Apply Lambert W
\\[
u = W\\left( -\\frac{\\ln 5}{k} \\right)
\\]
\\[
t = -\\frac{1}{\\ln 5} \\, W\\left( -\\frac{\\ln 5}{k} \\right)
\\]
\\[
x = t + \\frac{7}{4} = \\frac{7}{4} – \\frac{1}{\\ln 5} \\, W\\left( -\\frac{\\ln 5}{k} \\right)
\\]<\/p>\n\n\n\n
Recall \\( k = \\frac{100}{5^{7\/4}} \\), so:
\\[
-\\frac{\\ln 5}{k} = -\\frac{\\ln 5}{\\frac{100}{5^{7\/4}}} = -\\frac{\\ln 5 \\cdot 5^{7\/4}}{100}
\\]<\/p>\n\n\n\n
Thus:
\\[
x = \\frac{7}{4} – \\frac{1}{\\ln 5} \\, W\\left( -\\frac{\\ln 5 \\cdot 5^{7\/4}}{100} \\right)
\\]<\/p>\n\n\n\n
—<\/p>\n\n\n\n
6. Simplify the constant inside \\( W \\)
\\[
5^{7\/4} = 5^{1 + 3\/4} = 5 \\cdot 5^{3\/4}
\\]
But better: \\( 5^{7\/4} = 25 \\cdot 5^{-1\/4} \\) because \\( 5^{7\/4} = 5^{2 – 1\/4} = 25 \\cdot 5^{-1\/4} \\).<\/p>\n\n\n\n
Check: \\( 2 – 1\/4 = 7\/4 \\), yes.<\/p>\n\n\n\n
So:
\\[
\\frac{\\ln 5 \\cdot 5^{7\/4}}{100} = \\frac{\\ln 5 \\cdot 25 \\cdot 5^{-1\/4}}{100} = \\frac{\\ln 5}{4} \\cdot 5^{-1\/4}
\\]
Thus:
\\[
-\\frac{\\ln 5 \\cdot 5^{7\/4}}{100} = -\\frac{\\ln 5}{4 \\cdot 5^{1\/4}}
\\]<\/p>\n\n\n\n
Let \\( A = \\frac{\\ln 5}{4 \\cdot 5^{1\/4}} \\).<\/p>\n\n\n\n
Then:
\\[
x = \\frac{7}{4} – \\frac{1}{\\ln 5} \\, W(-A)
\\]<\/p>\n\n\n\n
—<\/p>\n\n\n\n
7. Find \\( W(-A) \\) for both real branches
We know \\( W(z) \\) is real for \\( z \\ge -1\/e \\).
Compute \\( A \\):
\\[
A = \\frac{\\ln 5}{4 \\cdot 5^{1\/4}}
\\]
Numerically: \\( \\ln 5 \\approx 1.609438 \\), \\( 5^{1\/4} \\approx 1.495349 \\), so:
\\[
A \\approx \\frac{1.609438}{4 \\times 1.495349} \\approx \\frac{1.609438}{5.981396} \\approx 0.269
\\]
So \\( -A \\approx -0.269 \\), which is \\( > -1\/e \\approx -0.3679 \\), so two real branches \\( W_0 \\) and \\( W_{-1} \\).<\/p>\n\n\n\n
—<\/p>\n\n\n\n
8. Solve for \\( W(-A) \\) exactly
We want \\( W(-A) \\) such that:
\\[
W(-A) e^{W(-A)} = -A = -\\frac{\\ln 5}{4 \\cdot 5^{1\/4}}
\\]<\/p>\n\n\n\n
First branch:
Try \\( W_0(-A) = -\\frac{\\ln 5}{4} \\).<\/p>\n\n\n\n
Check:
\\[
\\left(-\\frac{\\ln 5}{4}\\right) \\exp\\left(-\\frac{\\ln 5}{4}\\right) = -\\frac{\\ln 5}{4} \\cdot 5^{-1\/4} = -\\frac{\\ln 5}{4 \\cdot 5^{1\/4}} = -A
\\]
\u2713 Exact.<\/p>\n\n\n\n
So:
\\[
W_0(-A) = -\\frac{\\ln 5}{4}
\\]
\\[
x_1 = \\frac{7}{4} – \\frac{1}{\\ln 5} \\left( -\\frac{\\ln 5}{4} \\right) = \\frac{7}{4} + \\frac{1}{4} = 2
\\]<\/p>\n\n\n\n
—<\/p>\n\n\n\n
Second branch:
Try \\( W_{-1}(-A) = -\\frac{5}{4} \\ln 5 \\).<\/p>\n\n\n\n
Check:
\\[
\\left(-\\frac{5}{4} \\ln 5\\right) \\exp\\left(-\\frac{5}{4} \\ln 5\\right) = -\\frac{5}{4} \\ln 5 \\cdot 5^{-5\/4}
\\]
\\[
= -\\frac{\\ln 5}{4} \\cdot 5^{1} \\cdot 5^{-5\/4} = -\\frac{\\ln 5}{4} \\cdot 5^{-1\/4} = -\\frac{\\ln 5}{4 \\cdot 5^{1\/4}} = -A
\\]
\u2713 Exact.<\/p>\n\n\n\n
So:
\\[
W_{-1}(-A) = -\\frac{5}{4} \\ln 5
\\]
\\[
x_2 = \\frac{7}{4} – \\frac{1}{\\ln 5} \\left( -\\frac{5}{4} \\ln 5 \\right) = \\frac{7}{4} + \\frac{5}{4} = 3
\\]<\/p>\n\n\n\n
—<\/p>\n\n\n\n
9. Final solutions
\\[
x = 2 \\quad \\text{and} \\quad x = 3
\\]<\/p>\n\n\n\n
—<\/p>\n\n\n\n
\\[
\\boxed{2, 3}
\\]<\/p>\n\n\n\n
<\/p>\n\n\n\n
<\/p>\n","protected":false},"excerpt":{"rendered":"
Solve \\( 5^x = 100x – 175 \\) Alright \u2014 let\u2019s solve \\( 5^x = 100x – 175 \\) from scratch using the Lambert W method systematically, without guessing solutions. — 1. Rewrite the equationWe start with:\\[5^x = 100x – 175\\]\\[e^{x \\ln 5} = 100x – 175\\] — 2. Eliminate the constant term in the […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"ocean_post_layout":"","ocean_both_sidebars_style":"","ocean_both_sidebars_content_width":0,"ocean_both_sidebars_sidebars_width":0,"ocean_sidebar":"","ocean_second_sidebar":"","ocean_disable_margins":"enable","ocean_add_body_class":"","ocean_shortcode_before_top_bar":"","ocean_shortcode_after_top_bar":"","ocean_shortcode_before_header":"","ocean_shortcode_after_header":"","ocean_has_shortcode":"","ocean_shortcode_after_title":"","ocean_shortcode_before_footer_widgets":"","ocean_shortcode_after_footer_widgets":"","ocean_shortcode_before_footer_bottom":"","ocean_shortcode_after_footer_bottom":"","ocean_display_top_bar":"default","ocean_display_header":"default","ocean_header_style":"","ocean_center_header_left_menu":"","ocean_custom_header_template":"","ocean_custom_logo":0,"ocean_custom_retina_logo":0,"ocean_custom_logo_max_width":0,"ocean_custom_logo_tablet_max_width":0,"ocean_custom_logo_mobile_max_width":0,"ocean_custom_logo_max_height":0,"ocean_custom_logo_tablet_max_height":0,"ocean_custom_logo_mobile_max_height":0,"ocean_header_custom_menu":"","ocean_menu_typo_font_family":"","ocean_menu_typo_font_subset":"","ocean_menu_typo_font_size":0,"ocean_menu_typo_font_size_tablet":0,"ocean_menu_typo_font_size_mobile":0,"ocean_menu_typo_font_size_unit":"px","ocean_menu_typo_font_weight":"","ocean_menu_typo_font_weight_tablet":"","ocean_menu_typo_font_weight_mobile":"","ocean_menu_typo_transform":"","ocean_menu_typo_transform_tablet":"","ocean_menu_typo_transform_mobile":"","ocean_menu_typo_line_height":0,"ocean_menu_typo_line_height_tablet":0,"ocean_menu_typo_line_height_mobile":0,"ocean_menu_typo_line_height_unit":"","ocean_menu_typo_spacing":0,"ocean_menu_typo_spacing_tablet":0,"ocean_menu_typo_spacing_mobile":0,"ocean_menu_typo_spacing_unit":"","ocean_menu_link_color":"","ocean_menu_link_color_hover":"","ocean_menu_link_color_active":"","ocean_menu_link_background":"","ocean_menu_link_hover_background":"","ocean_menu_link_active_background":"","ocean_menu_social_links_bg":"","ocean_menu_social_hover_links_bg":"","ocean_menu_social_links_color":"","ocean_menu_social_hover_links_color":"","ocean_disable_title":"default","ocean_disable_heading":"default","ocean_post_title":"","ocean_post_subheading":"","ocean_post_title_style":"","ocean_post_title_background_color":"","ocean_post_title_background":0,"ocean_post_title_bg_image_position":"","ocean_post_title_bg_image_attachment":"","ocean_post_title_bg_image_repeat":"","ocean_post_title_bg_image_size":"","ocean_post_title_height":0,"ocean_post_title_bg_overlay":0.5,"ocean_post_title_bg_overlay_color":"","ocean_disable_breadcrumbs":"default","ocean_breadcrumbs_color":"","ocean_breadcrumbs_separator_color":"","ocean_breadcrumbs_links_color":"","ocean_breadcrumbs_links_hover_color":"","ocean_display_footer_widgets":"default","ocean_display_footer_bottom":"default","ocean_custom_footer_template":"","ocean_post_oembed":"","ocean_post_self_hosted_media":"","ocean_post_video_embed":"","ocean_link_format":"","ocean_link_format_target":"self","ocean_quote_format":"","ocean_quote_format_link":"post","ocean_gallery_link_images":"on","ocean_gallery_id":[],"footnotes":""},"categories":[1],"tags":[],"class_list":["post-82","post","type-post","status-publish","format-standard","hentry","category-uncategorized","entry"],"yoast_head":"\n