Question #3624b

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Question #3624b

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Answer 1 · 2017-06-13T17:38:13

$x = {log}_{\frac{25}{32}} 500$ $x=log_(25/32)500$

Explanation:

$5^{2 x - 3} - 2^{5 x + 2} = 0$ $5^(2x-3)-2^(5x+2)=0$
$5^{2 x - 3} = 2^{5 x + 2}$ $5^(2x-3)=2^(5x+2)$
Taking the natural logarithm of both sides,
$ln (5^{2 x - 3}) = ln (2^{5 x + 2})$ $ln(5^(2x-3))=ln(2^(5x+2))$
$(2 x - 3) ln 5 = (5 x + 2) ln 2$ $(2x-3)ln5=(5x+2)ln2$
$(2 ln 5) x - 3 ln 5 = (5 ln 2) x + 2 ln 2$ $(2ln5)x-3ln5=(5ln2)x+2ln2$
$(2 ln 5) x - (5 ln 2) x = 3 ln 5 + 2 ln 2$ $(2ln5)x-(5ln2)x=3ln5+2ln2$
$[ln (5^{2}) - ln (2^{5})] x = ln (5^{3}) + ln (2^{2})$ $[ln(5^2)-ln(2^5)]x=ln(5^3)+ln(2^2)$
$x = \frac{ln (5^{3} \cdot 2^{2})}{ln (\frac{5^{2}}{2^{5}})}$ $x=ln(5^3*2^2)/ln(5^2/2^5)$
$x = \frac{ln (125 \cdot 4)}{ln (\frac{25}{32})}$ $x=ln(125*4)/ln(25/32)$
$x = {log}_{\frac{25}{32}} 500$ $x=log_(25/32)500$

Answer 2 · 2017-06-13T17:53:05

Given: $5^{2 x - 3} - 2^{5 x + 2} = 0$ $5^(2x-3)-2^(5x+2)= 0$

Move the second term to the right:

$5^{2 x - 3} = 2^{5 x + 2}$ $5^(2x-3)=2^(5x+2)$

Use the base 5 logarithm on both sides:

${log}_{5} (5^{2 x - 3}) = {log}_{5} (2^{5 x + 2})$ $log_5(5^(2x-3))=log_5(2^(5x+2))$

Use the identity ${log}_{b} (a^{c}) = (c) {log}_{b} (a)$ $log_b(a^c) = (c)log_b(a)$ on both sides:

$(2 x - 3) {log}_{5} (5) = (5 x + 2) {log}_{5} (2)$ $(2x-3)log_5(5)=(5x+2)log_5(2)$

Use the property ${log}_{b} (b) = 1$ $log_b(b) = 1$ on the left:

$2 x - 3 = (5 x + 2) {log}_{5} (2)$ $2x-3=(5x+2)log_5(2)$

Use the distributive property on the right:

$2 x - 3 = 5 {log}_{5} (2) x + 2 {log}_{5} (2)$ $2x-3=5log_5(2)x+2log_5(2)$

Subtract $5 {log}_{5} (2) x$ $5log_5(2)x$ from both sides:

$(2 - 5 {log}_{5} (2)) x - 3 = 2 {log}_{5} (2)$ $(2-5log_5(2))x-3=2log_5(2)$

Add 3 to both sides:

$(2 - 5 {log}_{5} (2)) x = 3 + 2 {log}_{5} (2)$ $(2-5log_5(2))x=3+2log_5(2)$

Divide both sides by the coefficient of x:

$x = \frac{3 + 2 {log}_{5} (2)}{2 - 5 {log}_{5} (2)}$ $x=(3+2log_5(2))/(2-5log_5(2))$

Convert to base e by using the conversion formula ${log}_{5} (x) = \frac{ln (x)}{ln (5)}$ $log_5(x)= ln(x)/ln(5)$ :

$x = \frac{3 + 2 \frac{ln (2)}{ln (5)}}{2 - 5 \frac{ln (2)}{ln (5)}}$ $x=(3+2ln(2)/ln(5))/(2-5ln(2)/ln(5))$

Multiply by 1 in the form of $\frac{ln (5)}{ln (5)}$ $ln(5)/ln(5)$ :

$x = \frac{3 ln (5) + 2 ln (2)}{2 ln (5) - 5 ln (2)}$ $x=(3ln(5)+2ln(2))/(2ln(5)-5ln(2))$

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Question #3624b

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